Problem Solving
Through
Recreational
Mathematics
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Bonnie Averbach
AND
Orin Chein
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Problem Solving
Through
Recreational Mathematics
Bonnie Averbach
and
Grin Chein
DOVER PUBLICATIONS, INC.
Mineola, New York

Copyright
Copyright © 1980 by W H. Freeman and Company
All rights reserved under Pan American and International Copyright
Conventions.
Published in Canada by General Publishing Company, Ltd., 30 Lesmill Road,
Don Mills, Toronto, Ontario.
Bibliographical Note
This Dover edition, first published in 2000, reproduces entirely the text of
Mathematics: Problem Solving Through Recreational Mathematics, originally
published in 1980 by W. H. Freeman and Company, San Francisco. A
substantial new appendix on the subject of probability, with accompanying exercises,
hints and solutions, and answers, as well as a publisher's note have been added
to the present edition. In addition, the index, contents, and some of the
prefatory material have been updated to incorporate the new material.
Library of Congress Cataloging-in-Publication Data
Averbach, Bonnie, 1993-
[Mathematics]
Problem solving through recreational mathematics / Bonnie Averbach and
Orin Chein.
p. cm.
Originally published: Mathematics. San Francisco : W. H. Freeman, cl980.
Includes bibliographical references and index.
ISBN 0-486-40917-1 (pbk.)
1. Mathematics. 2. Mathematical recreations. 3. Problem solving. I. Chein,
Orin, 1943- II. Title.
QA39.2 .A896 2000
510—<ic21
99-052198
Manufactured in the United States of America
Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

Contents
•^ Indicates sections that may be omitted without loss of continuity.
Preface ix
To the Reader xiii
Acknowledgnnents
1 Following the Clues
XV
1
Sample Problems 2
Which Chart or Diagram to Choose
Presenting a Solution 11
Some Steps in Problem Solving 11
Tree Diagrams 16
The Multiplication Principle
Simplification 22
The Chapter in Retrospect
Exercises 26
19
25
2 Solve It With Logic 37
Sample Problems 38
Statements 39
Variables and Connectives 41
Negation 41
'And' - Conjunction 42
'Or-Disjunction 43
Conditional and Biconditional
Statements 44
Drawing Conclusions 47
Compound Statements 49
lO
Logical Implication and Equivalence 53
Arguments and Validity 55
The Chapter in Retrospect 60
Exercises 61
From Words to Equations:
Algebraic Recreations 70
Sample Problems 71
Introducing Variables 72
The Chapter in Retrospect 84
Exercises 84
Solve It With Integers: Some Topics
From Number Theory 100
Sample Problems lOl
Diophantine Equations 103
Divisibility 104
Prime Numbers 106
^ The Infinitude of Primes 106
^ The Sieve of Eratosthenes 107
More About Primes 108
Linear Diophantine Equations 115
Division With Remainders 119
Congruence 120
Casting Out Nines 125
Solving Linear Congruences 127

CONTENTS
145
Solving Linear Diophontine Equations 131
The Chapter in Retrospect 137
Exercises 138
More About Numbers:
Bases and Cryptarithmetic
Sample Problems 146
Positional Notation 147
Changing Bases 148
Addition and Multiplication in Other
Bases 153
Cryptarithmetic 156
The Chapter in Retrospect 162
Exercises 163
Solve It With Networks: An
Introduction to Graph
Theory 173
Sample Problems 174
Graphs 175
Eulerian Paths and Circuits 178
Odd and Even Vertices 180
^ More Than Two Odd Vertices 187
^ Directed Graphs 190
Hamiltonian Circuits 192
The Knighfs Tour 194
other Applications 196
^ Coloring Graphs and Maps 198
The Chapter in Retrospect 201
Exercises 202
Games of Strategy for Two
Players 213
Sample Problems 214
Chance-Free Decisionmaking 215
Games of Perfect Information 216
Finiteness 216
The Existence of Winning Strategies 217
Position-State of the Game 224
The State Diagram of a Game 228
How Do We Find a Winning Strategy? 230
Finding a Winning Strategy by Working
Backward 230
Finding Winning Strategies by Simplifying a
Game 232
Finding Winning Strategies With a Frontal
Assault 233
How Many Possibilities Need Be
Considered? 234
Symmetry as a Limiting Factor 234
D^jd Vu-We've Seen It Before 241
The Game of Nim 246
Pairing Strategies 250
Variations of a Game 251
The Chapter in Retrospect 252
Exercises 253
8 Solitaire Games and
Puzzles 273
Sample Problems 274
The Tower of Brahma 276
Dissection Problems 279
Polyominoes 281
Soma 284
Peg Solitaire 285
The Fifteen Puzzle 288
Even and Odd Permutations 291
^ Coloring and the 15 Puzzle-A Second
Approach 298
Colored Cubes 299
^ Colored Cubes-A Second
Approach 303
The Chapter in Retrospect 304
Exercises 305
9 Potpourri 313
Decimation 313
Coin Weighing 315
Shunting 316
Syllogisms 317
Grab Bag 318
The Book in Retrospect 320

CONTENTS
Appendix A—Some Basic
Algebraic Techniques 321
Appendix B—Mathematical
Induction 332
Appendix C—Probability 337
Bibliography 376
Hints and Solutions
Answers to Selected
Problems 433
Index 455
380

Publisher's Note
This book is a reproduction of Mathematics: Problem Solving through
Recreational Mathematics which was published by W. H. Freeman and
Company in 1980. The original manuscript for that book contained a
chapter on probability which was omitted from that edition. We still
believe that the material in this chapter (which was Chapter 7 in the
original manuscript) is worthy of publication, so in this new Dover
edition it has been reinstated as Appendix C, found on page 336. This
appendix includes all the detailed discussion, sample problems,
exercises, hints and solutions, and selected answers that any of this book's other
chapters include; we feel it makes the present edition an even greater aid
to students and instructors than the original edition, which is now
unavailable.
When Mathematics: Problem Solving through Recreational Mathematics
was in print, it was accompanied by a solutions manual available to
instructors using the book as a text for their courses.This manual is now
available from the authors, at a nominal charge. Readers desiring a copy
of the manual should contact Bonnie Averbach at the Department
of Risk Management and Actuarial Science, Temple University,
Philadelphia, PA 19122.

Preface
How can students be given a feeling for mathematics in a one or
two semester college course? With most of the usual approaches—
through history, culture, or applications—the student is an observer.
He or she learns about what others have done, but does not learn to
think independently. The student learns some techniques and practices
them on problems that are essentially the same as those done in class.
If applications are stressed, only highly simplified cases can usually
be considered. For example, linear programming techniques are usually
presented for functions of only two variables, a case that rarely comes
up in practice. The result is that the student comes away with little
of permanent benefit and unconvinced of the actual usefulness of the
subject.
A few years ago, we decided to try a different approach. We said to
the student: "You participate and be the mathematician. Take a
problem and use anything and everything you know to solve it. Think
about it; strain your mind and imagination; put it aside, if necessary;
keep it in mind; come back to it. If you can solve it on your own,
isn't the feeling great? If you can't solve it, maybe some mathematics
(new to you) would be helpful to know; let's develop some and see."
What have we accomplished with this approach? As teachers we
love teaching this course, because of the enthusiasm generated by the
students. The students we taught learned that mathematics is not just
numbers and manipulation—it's thinking; it's strategy. In fact, aspects
of mathematics can be seen all around us and are useful for any kind
of problem, in any kind of setting. Most important, our students have
learned how to think critically. Several of our former students have
returned to tell us how they have been able to apply techniques that
they learned in our course to a variety of situations, ranging from
student teaching to business.

PREFACE
Our approach is based on recreational mathematics—recreational
problems, puzzles, and games. Why? First, because they are fun, and
motivation is perhaps the central problem in teaching mathematics.
Second, because, historically, many important mathematical concepts
arose from problems that were recreational in origin. Recreational
mathematics has proved ideal for introducing most of the topics usually
covered in liberal arts mathematics courses. However, our emphasis
is not on the mathematical results themselves, but rather on how
these results can be used in thinking about problems and solving them.
Problem solving is one of the most important skills a person can
acquire. It is useful to the mathematician and is also of great
importance in many other fields of endeavor and in life itself.
The major difficulty we encountered in developing a course based
on recreational problem solving was the selection of a text. Although
there are many books available that contain excellent selections of
mathematical problems, puzzles, and games, none was completely
suitable for our purposes. Very few of these books attempt to explain
how to go about attacking a problem or try to present any formal
mathematics. For this reason, we decided to write our own notes for
the course—the basis for this book.
During the several years in which we have been teaching this course,
the actual topics covered have varied from semester to semester. Since
we have decided to include most of these topics in this book, there is
more than enough material for a one semester course. Because of this
abundance of material, an instructor using this book as a text for a
course lasting one semester or less must select which chapters to cover.
For this reason, we have made the chapters as independent of each
other as possible.
The only dependencies are the following:
1. The ability to translate a verbal quantitative problem into
equations is required for some of the problems and exercises in
Chapter 4. This translation process is discussed briefly in Chapter 3,
and drill in using it can be obtained from the Chapter 3 exercises.
A student with a good background in algebra could omit Chapter 3
completely.
2. Some of the exercises in Chapter 5 may be solved most simply
by applying the notion of congruence, which is discussed in Chapter 4.
These exercises are designated by the symbol #.
3. Some of the games in Chapters 7 and 8 refer back to ideas
presented in earlier chapters. In particular, a matchstick game is first
introduced in Chapter 1; congruences (Chapter 4) are useful in
analyzing some of the Chapter 7 matchstick games; the binary system
(Chapter 5) is needed for a discussion of Nim (7) and is also of
interest with regard to the Tower of Brahma (8); the two solutions

PREFACE
of the colored cubes problem (8) refer to unique factorization (4)
and graph theory (6).
The chapters are otherwise independent, although we do recommend
beginning with Chapter 1.
Each chapter of the book contains several types of problems with
different purposes. The Sample Problems are to whet the reader's
appetite and motivate the discussions of the chapter (these are solved
in the text). The Practice Problems are to firm up the reader's grasp
of the mathematical ideas presented. The Exercises are to challenge
the reader and help develop problem solving ability. Exercises that
are unusually difficult or lengthy are marked by M (or, if they are
very difficult, by '^M). The notation [a] after an Exercise or Practice
Problem number indicates that the answer to the problem is given in
the Answers to Selected Problems in the back of the book; [h] indicates
that a hint is given in the Hints and Solutions; and [s] indicates that
the problem is solved completely in the Hints and Solutions. In each
chapter, a number of exercises are solved completely. These solutions
augment the discussion of the chapter and occasionally present some
techniques that will be helpful with other problems.
The sample problems and exercises in this book are, for the most
part, problems we have accumulated over the years. These are mostly
our variations of problems given us by students and colleagues or that
have appeared in columns, journals, and the many puzzle books in
recreational mathematics. Some of these problems may be found, in
some form or another, in more than one source; many are now
considered to be classics. As a result, although we originally intended
to trace the true origin of each problem, we found this a difficult, if
not impossible, task. For this reason and with a few exceptions, we
decided to give references only for problems that we have taken
essentially verbatim. However, we want to draw your attention to the
many excellent books on recreational mathematics. These have been
an inspiration to us and to our students, and we hope they will
inspire others as well. A selected list of these may be found in the
Bibliography on page 376. In addition, when we are aware of the
creator of a problem or game, we have so indicated in parentheses.
However, our knowledge in this matter is very incomplete, and we
apologize for any mistakes or oversights.
May 1980 Bonnie Averbach
Orin Chein

To the Reader
Problem solving can be a stimulating and exciting pastime. Sometimes
one can see the method of solution immediately. At other times, a
problem can lead to hours of frustration. But the satisfaction that
comes with the eventual solution of a difficult problem is, in itself, a
reward that makes all the effort worthwhile.
In working with this book, think of yourself as the mathematician,
the problem solver. Near the beginning of each chapter, you will find
some sample problems. Try to solve the first of these before reading
the chapter. If you are successful, think about how you approached
the problem—specifically what you did to solve it. If you are
unsuccessful, think about what further information would be helpful to
you, what kind of mathematical considerations might be useful.
As you read the chapter, keep the original sample problem in mind.
If you reach a point at which you think you might be able to solve the
original problem, stop reading and try it again. Then read on to see our
discussion of the solution. (If you have been able to solve the problem,
you may find that your solution is better—simpler or more elegant —
than ours. However, continue reading our discussion anyway. We have
usually chosen our approach for a reason, and, in any case, it is often
instructive to see different approaches to the same problem.)
After you have solved (or read about the solution of) the first
sample problem, try the second one. If you get stuck, read on. Continue
this approach throughout the chapter.
Do not be discouraged if you cannot solve all, or even any, of the
sample problems in a given chapter. Each problem is there to try to lead
you to an idea or a mathematical area that may be new to you. Even
when you are unable to solve a problem, if you have really thought
about it, then you have learned something. You may have partial
success. You may realize where the difficulties lie. But, even if you

TO THE READER
made no progress at all with the problem, when the solution is finally
shown to youj you will better appreciate the power of the techniques
and mathematical theory used.
When you have finished reading a chapter, try as many of the
exercises as you can. We do not expect you to attempt all the problems
immediately (leave some for your future enjoyment), but you only learn
to be a good problem solver by solving problems.
Although most of the exercises in each chapter relate to the basic
mathematical tools or strategies presented in the chapter, you will find
that being able to solve one problem does not mean that you will be able
to solve them all—each problem presents its own challenge.
Analyze each exercise you try, as you did the sample problems. If
you have been successful with a problem, try to take note of your
approach. What were the crucial steps? If you have been unsuccessful
after a reasonable effort, don't give up. Put the problem aside for a
while, but keep it in mind; sleep on it. Good ideas often come at the
strangest times.
If you are reading this book as part of a group or class, become
involved in any class discussions. Active participation is far more
beneficial than passive attendance. Ask questions; create your own
problems; share your ideas with others; try to provide helpful
criticism of the thought processes of others; and seek criticism of your
own thinking.
The more you get involved with the problems in this book, the
more benefits you will derive from the book. If you approach the
problems conscientiously, we think you will find that you will not only
become a better problem solver, but your general thinking processes,
as they apply to all fields, will benefit from the experience.
B.A.
O.C.

Acknowledgments
Like most authors, we are indebted to many people who have
assisted us in a variety of ways from the time our idea for this book
was first conceived until final publication.
For their advice, encouragement, and many helpful suggestions, we
would like to thank David Blauer, Isidor Chein, Martin Gardner,
Nancy Hagelgans, William Karjane, Archibald Krenzel, Robert Morse,
Leon Steinberg, William Traylor, William Wisdom, and David
Zitarelli, as well as the reviewers who read our manuscript at various
stages during its preparation and the editors who worked with us
during the production process.
We would also like to thank Raelaine Ballou for her continual
assistance in tracking down some of the many sources we used in
preparing our manuscript, Susan Graham for drawing preliminary
sketches for many of our cartopn ideas, John Johnson for embellishing
these ideas and bringing them to their delightful final form, and
Nadia Kravchenko for her excellent typing of the entire manuscript.
For their constant encouragement and patience throughout the years
we were writing this book, we extend our love and thanks to the
members of our families—RacheUe, Wayne, Debra, Gary, Robert,
Stephen, Jake, and Alex Averbach, and Carrie, Adam, and Jason Chein.
The origins of many mathematical problems are lost in the mists of
antiquity; thus our attempts to locate original sources for our problems
were not very successful. We decided to settle for identifying those
sources from which we have taken problems essentially verbatim. Each
such problem in the text is followed by a reference number, which
refers to a source appearing in the Bibliography. In particular, we
would like to acknowledge the following publishers and authors for
granting us permission to reprint or adapt some of their material.
Exercises 1.19 and 1.22 are reprinted with the permission of

ACKNOWLEDGMENTS
Al B. Perlman. Problem 4.4 and Exercises 5.9 and 7.22 have been
adapted from Mathematical Puzzles for Beginners and Enthusiasts by
Geoffrey Mott-Smith (copyright © 1946 by Copeland and Lamm, Inc.,
copyright renewed in 1974; published by Dover Publications, Inc.),
with the permission of Copeland and Lamm, Inc. Problem 8.2 and
its solution and Exercise 8.9 have been adapted from Polyominoes by
Solomon W. Golomb (copyright © 1965 by Solomon W. Golomb),
with the permission of Charles Scribner's Sons and Allen & Unwin,
Ltd.
Dover Publications has granted us permission to use problems from
several of the books in their fine series on mathematical recreations. In
particular, we have taken problems from the Dover books by Friedland,
Kraitchik, Phillips, and Read listed in our Bibliography, and also from
Mathematical Bafflers by Angela Dunn, some of whose material was
published earlier by Litton Industries in their Problematical Recreations
series.
Several journals have also been helpful: Technology Review, with
its "Puzzle Comer" edited by Allan Gottlieb; Arithmetic Teacher^
Mathematics Magazine}, and Sphinx: Revue Mensuelle des Questions
Recreatives, edited by M. Kraitchik and M. Pigeolet.
The American Mathematical Society, Emerson Books, Inc., the
Longman Group Limited, and McGraw-Hill, Inc. also granted
permission for use of problems.
Finally, we wish to thank Phillip Matthew Swift for his work
in graphics and word processing to prepare the new chapter on
probability (Appendix C) for this edition.
B.A.
O.C.

Following the Clues
An enjoyable and interesting way of studying problem solving in
general is to consider recreational problems in particular.
Recreational problems and games have been a source of amusement
and interest for hundreds of years. Although the origins of many
problems can be traced back to ancient times, probably the first
important collection of problems is a Greek Anthology attributed to
Metrodorus (c. 500). Many other collections followed. Among these,
Problemes plaisans et delectables, qui se font par les nombres, by Claude-
Gaspar Bachet de Meziriac (1612), is the first noteworthy collection
to appear in print.
It was not until the late nineteenth and early twentieth centuries,
though, that recreational mathematics acquired great popularity. Not
only were many books on the subject published at that time, but the
newspaper columns of Sam Loyd in America and H. E. Dudeney in
England helped to bring recreational puzzles to the attention of the
general public. The decades since have seen a continuation of this
trend: The newspaper columns of Hubert Phillips ("Caliban") in
England, the Belgian journal Sphinx: Revenue mensuelle des questions
recreativeSy the American Mathematical Monthly^ and many books
continue to supply a wide selection of recreational problems.
Current developments may be followed in ih^ Journal of Recreational
Mathematics, in the books of Martin Gardner, and in his columns in
Scientific American.
Although some mathematical background is required to solve certain
problems, many problems of recreational mathematics can be solved

CHAPTER 1
with little more than fundamental reasoning skills, a power of
concentration, perseverance, and a touch of imagination and ingenuity.
The techniques of solution can be used to help solve problems in
all fields.
We will investigate some of these techniques by considering the
following problems.
SAMPLE
PROBLEMS
Problem 1.1
An elimination boxing tournament was organized. There were 114
participants and so there were 57 matches in the first round of the
tournament. In the second round, the 57 fighters remaining were
paired, resulting in 28 matches; one fighter received a bye (that is,
did not have to fight in that round). The 29 fighters remaining were
then paired, and so on.
(a) How many matches in all were required to determine a winner
of the tournament?
(b) How many matches would be required if n people participated
in the tournament (where n represents a fixed but unspecified whole
number) ?
Problem 1.2
Ms. X, Ms. Y, and Ms. Z—an American woman, an Englishwoman,
and a Frenchwoman, but not necessarily in that order, were seated
around a circular table, playing a game of Hearts. Each passed three
cards to the person on her right. Ms. Y passed three hearts to the
American. Ms. X passed the queen of spades and two diamonds to
the person who passed her cards to the Frenchwoman.
Who was the American? The Englishwoman? The Frenchwoman?
Problem 1.3
Armand Alloway, Basil Bennington, Col. Carlton Cunningham,
Durwood Dunstan, and Everett Elmsby, Esq. are the five senior
members of the Devonshire Polo Club. Each owns a pony that is
named after the wife of one of the others.
Mr. AUoway's pony is named Georgette; Col. Cunningham owns
Jasmine; and Mr. Elmsby owns Inez. Francine, owned by Mr.
Dunstan, is named after Alloway's wife. Georgette's husband owns
the pony that is named after Mr. Bennington's wife. Helene
Cunningham is the only wife who knows how to ride a horse.
Who is Jasmine's husband? Who owns Helene?

FOLLOWING THE CLUES
Problem 1.4
Messrs. Baker, Dyer, Farmer, Glover, and Hosier are seated around a
circular table, playing poker. Each gentleman is the namesake of the
profession of one of the others. The dyer is seated two places to the
left of Mr. Hosier. The baker sits two places to Mr. Baker's right. The
farmer is seated to the left of Mr. Farmer. Mr Dyer is on the glover's
right.
What is the name of the dyer? ([51],* Problem 36)
Problem 1.5
Two players, A and B, take turns in the following game. There is a
pile of six matchsticks. At a turn, a player must take one or two sticks
from the remaining pile. The player who takes the last stick wins.
Player A makes the first move and each player always makes the best
possible move.
Who wins this game?
Problem 1.6
Figure 1.1 pictures a railroad track on which are found a locomotive
and two railroad cars. Car B at the right has just been filled with
coal from the conveyor belt; car A at the left is empty. The tunnel
is large enough to accommodate either car but not the locomotive, L.
Furthermore, each car is longer than the tunnel so that, when in the
tunnel, it is accessible to the locomotive from either side.
Using the locomotive to push or pull the cars, how can the two
cars be made to switch places and the locomotive end up between them ?
tunnel
FIGURE 1.1
* References appear in the bibliography at the end of the book (page 336).

CHAPTER 1
Before reading further, try to solve Problem 1.1 (page 2).
Solution of
Problem 1.1
Did you approach part (a) of the problem as follows, by counting
matches ?
1st round: 57 matches, 57 fighters remain.
2nd round: 28 matches, 29 fighters remain.
3rd round: 14 matches, 15 fighters remain.
4th round: 7 matches, 8 fighters remain.
5th round: 4 matches, 4 fighters remain.
6th round: 2 matches, 2 fighters remain.
7th round: 1 match, 1 winner remains.
Thus, 57-}-28-}-14-}-7-}-4-}-2-}-l = 113 matches must be held.
If you used this approach, you have found a correct solution for
part (a).
Does this method carry over to give a solution for part (b)? Not
readily.
Is there another way in which you can approach the problem? Can
you adopt another point of view?
Consider part (a) again. Since there could be only one winner, 113
fighters had to be eliminated; to eliminate 113 fighters requires 113
matches.
Although this second approach seems simpler, both methods of
solving part (a) are correct. The advantage of the second approach
is that it does carry over to part (b): Since n - 1 fighters must be
eliminated, w — 1 matches are required.
What can we learn from this example? Often, there is more than
one way to look at a problem. Sometimes several ways will lead to a
solution; other times, only one. Even when several approaches to a
problem do succeed, one approach may be simpler or more satisfying
than another. (As you gain more experience in problem solving, you
may find several different solutions to a problem. You can then compare
these and select the one that seems the most elegant to you.)
In this sense, solving a problem is like finding the way through a
maze or labyrinth. There may be several paths that lead to the goal,
but one path may be shorter than the others.
On the other hand, there may be only one correct path and the
other paths may lead to blind alleys. Similarly, many approaches to a
particular problem may lead to a dead end.
At the entrance of a labyrinth, there is usually no way to know
which is the correct path to take. We try one and, if we reach a blind

FOLLOWING THE CLUES
alley or seem to be going around in circles, we return to the
beginning and try another.* Similarly, there are no hard and fast rules
that tell us how to approach every problem. If we seem to be getting
nowhere with one approach, then we try another. Since there are only
a finite number of possible paths in a labyrinth, eventually, by trial
and error, we will hit upon the right one. Unfortunately, however,
not all the paths in solving a problem are clearly visible. Sometimes
they are, but often success as a problem solver depends on the ability
to discover the "hidden paths."
Fortunately, however, there are important aspects of solving
problems that don't apply in traversing a labyrinth. When we are in
a labyrinth, we can see only the paths immediately before us but have
no idea where they will lead. When we attack a problem, on the other
hand, we usually have an overview of the entire arena—that is, we
can look at the problem from above rather than within. This overview
can give us different perspectives from which we can gain added insight.
In addition, we may be able to draw on our experience. We may
recognize similarities with other problems. If one problem reminds
us of another that we have already solved, we may try the techniques
of the original solution as our first approach to the new problem.
In this sense, the more experience we have in problem solving, the more
successful we are likely to be in attacking new problems.
Although there are no hard and fast rules applicable to all problems,
there are some basic techniques for starting to find a solution. Before
we discuss some of these techniques, you should attempt some of the
sample problems again. As you work on them, try to be aware of your
thinking processes. What techniques prove successful? Where do you
encounter difficulties?
Try Problems 1.2 and 1.3 (page 2).
Now that you have thought about and possibly solved these
problems, we will use them as models to illustrate some techniques.
In Problem 1.2, we are given the following information:
1. Each person passed to her right.
2. Ms. Y passed to the American. (The fact that she passed three
hearts is irrelevant.)
* There are systematic ways of attacking a labyrinth. See, for example, [13], pp. 127-137.
Solution of
Problem 1.2

CHAPTER 1
3. Ms. X passed to the person who passed to the Frenchwoman.
How can we use this information?
A diagram will help. Since Ms. Y passed to the American, she
is not the person who passed to the Frenchwoman (Figure 1.2). Ms.
X passed to the person who passed to the Frenchwoman (Figure 1.3).
Frenchwoman
Ms. Y
American
(not Frenchwoman)
Ms.X
FIGURE 1.2
FIGURE 1.3
Did Ms. X pass to Ms. Y? Clearly not, when you compare Figures
1.2 and 1.3.
Therefore, Ms. X passed to Ms. Z (Figure 1.4), and Ms. Y is the
Frenchwoman (Figure 1.5).
Frenchwoman
Ms. X V y Ms. Z
FIGURE 1.4
Ms.X
FIGURE 1.5
Frenchwoman
Ms.Y
O
Ms. Z
Combining these with Figure 1.2, we find that Ms. X is the
American (Figure 1.6), which leaves Ms. Z to be the Englishwoman
(Figure 1.7).
Frenchwoman
Ms.Y
American
Ms.X
Ms. Z
Frenchwoman
Ms.Y
American V J Englishwoman
Ms.X N / Ms. Z
FIGURE 1.6
FIGURE 1.7

FOLLOWING THE CLUES
The solution of this problem was relatively easy. All we had to do
was list the pertinent information that was given and draw some
straightforward conclusions. With the aid of several diagrams, out came
the desired identifications.
Before we try to make a list of the steps we followed in attacking
this problem, let us consider Problem 1.3 (page 2).
We begin by listing the information given in the statement of the
problem:
1. Mr. Alloway's pony is Georgette.
2. Col. Cunningham's pony is Jasmine.
3. Mr. Elmsby's pony is Inez.
4. Mr. Dunstan's pony is Francine.
5. Francine is married to Mr. Alio way.
6. Helene is married to Col. Cunningham.
7. Georgette's husband owns the pony that is named after Mr.
Bennington's wife.
8. Each man's pony is named after somebody else's wife, and not
after his own wife.
Here, a chart with three columns is helpful—one column each for
the name of the gentleman, the name of his horse, and the name
of his wife. By entering the given information along with the number
of the pertinent clue, we obtain Figure 1.8.
Solution of
Problem 1.3
gentleman
Mr. Alloway
Col. Cunningham
Mr. Elmsby
Mr. Dunstan
Mr.X
Mr. Bennington
horse
Georgette (1)
Jasmine (2)
Inez (3)
Francine (4)
Y
wife
Francine (5)
Helene (6)
Georgette (7)
Y (7)
FIGURE 1.8

CHAPTER 1
Note that the only way in which clue 7 could be entered in the
chart was to introduce symbols such as X and Y to represent people
who have not yet been identified. (This is similar to the technique of
algebra in which letters are used to represent unknown number
quantities.) In Figure 1.8, Mr. X represents Georgette's husband and
Y represents the name of Mr. Bennington's wife, which is also the
name of Mr. X's horse.
We now make several observations:
Mr. Bennington's horse must be Helene because the other four
names are already assigned in the horse column.
Mr. X cannot be Col. Cunningham, as the latter is married to
Helene whereas Mr. X is married to Georgette.
Hence, since Col. Cunningham owns Jasmine and Mr. X owns Y,
Y cannot be Jasmine.
Nor can Y be Francine or Helene, since Y is married to Mr.
Bennington, and Francine and Helene are not.
Y cannot be Georgette, since Georgette's husband owns Y, according
to clue 8.
The only possibility remaining is that Y is Inez. But then Mr. X
must be Mr. Elmsby (since he owns Inez).
We can now consolidate our chart to obtain Figure 1.9. By the
process of elimination, we have discovered that Jasmine is married to
Durwood Dunstan.
gentleman
Mr. Alloway
Col. Cunningham
Mr. Elmsby
Mr. Dunstan
Mr. Bennington
horse
Georgette
Jasmine
Inez
Francine
Hdene
wife
Francine
Helene
Georgette
Jasmine
Inez
FIGURE 1.9
Alternative
Approach to
Problem 1.3
Instead of the chart used above, we could have used a table of the
type shown in Figure 1.10. The letters A, B, C, D, E are the first
letters of the gentlemen's last names, and the letters F, G, H, I, J are
the initials of the wives' first names. When we discover that a particular
gentleman is married to a particular lady, we mark V where his row
meets her column; if we know they are not married, we place an X.
Once a check appears, then the other boxes of that row and column
must contain X's, as each man is married to only one woman and

FOLLOWING THE CLUES
each woman is married to only one man. With this in mind, we can
deduce Figure L11 from clues 5 and 6.
Also, from clue 3, we know that E is not married to I and, from
clue 7, that B is not married to G. (Otherwise, Mr. B's pony would
be named after Mr. B's wife, contrary to clue 8.) We obtain Figure 1.12.
But now we seem to be stuck. It appears that we have not made full
use of clue 7; but how can we use it?
Since, in Figure 1.12, there are only two possible alternatives for
Georgette's husband, let us see what happens when we assume one of
them to be true.
Let us first marry Georgette off to Mr. Dunstan. Then, from clue 7,
Mr. Dunstan owns the pony that is named after Mr. Bennington's
wife. But we know that Mr. Dunstan's pony is Francine (clue 4) and
that Francine is not Mr. Bennington's wife (clue 5). We thus reach a
contradiction, and so this marriage does not work, and perhaps never
could have:
Georgette is not married to Mr. Dunstan.
Note that when an assumption logically leads to a contradiction (that is,
something is shown to be true that has previously been seen to be false,
or vice versa), then the assumption must be incorrect. The item assumed to
be true must therefore be false.
We enter this new information in our chart, and get Figure 1.13.
Observe that there is now only one possibility for Georgette's
husband—Mr. Elmsby.
We may, therefore, put a check in the E-G box in the chart and
then put an X in the E-J box.
Using clue 7 again, Mr. Elmsby owns the pony named after Mr.
Bennington's wife. But Mr. Elmsby owns Inez. It follows that:
Mr. Bennington is married to Inez.
Entering this information in the chart, we arrive at Figure 1.14.
We now put X's in the B-J and D-I boxes; and, hence, the D-J
box must receive a check.
Thus, as we found before.
Jasmine is married to Mr. Dunstan.
A
B
C
D
E
F
G
H
I
J
FIGURE 1.10
A
B
C
D
E
F
y
X
X
X
X
G
X
X
H
X
X
y
X
X
I
X
X
J
X
X
FIGURE 1.11
A
B
C
D
K
F
y
X
X
X
X
G
X
X
X
H
X
X
X
X
I
X
X
X
J
X
X
FIGURE 1.12
A
B
C
D
H
y
X
V
X
X
Ci
X
X
X
X
H
X
X
X
-
I
X
X
X
J
X
A
B
C
I)
K
F
y
X
X
X
X
G
X
X
X
X
H
X
X
v/
X
X
I
J
X 1 X
X X
X
"xj
FIGURE 1.13
FIGURE 1.14

10
CHAPTER 1
As a final step in the solution, regardless of which approach was used,
the original problem should be checked to verify that the conclusions
reached are consistent with the given facts.
WHICH CHART OR DIAGRAM TO CHOOSE
In our two approaches to the solution of Problem 1.3, a specific chart
or table helped us to envision the solution. There are other useful
visual aids. For example, in a table we could head the rows with the
horses' names and the columns with the women's names. We would
then place a check in, say, the F-row, J-column if Francine is the
name of the horse that belongs to the man who is married to Jasmine.
Or, we could plot the men against their horses; this time, instead of
checks, we use circles with the wives' initials, as in Figure 1.15.
This figure tells us that Mr. Alloway is married to Francine and
owns Georgette, Mr. Bennington does not own Inez, and Mr. Dunstan
owns Francine (but we do not yet know to whom he is married).
Another diagram we could use is Figure 1.16.
A
B
C
D
E
F
O
G
©
H
I
X
J
FIGURE 1.15
FIGURE 1.16
Here the letters G, H, J, F, I stand for both the wives' and the
horses' initials. We indicate that a particular horse belongs to a
particular owner by writing the horse's initial below that of its owner.
The lines connect husbands' and wives' initials.
In general, it is difficult, if not impossible, to know in advance
exactly what type of visual aid will be most helpful. Sometimes, none is
needed; other times, more than one is necessary. You might start out
with one aid and realize that another is better. The best guideline
seems to be to try to use the chart or other aid that will incorporate
as much of the given information as possible, in the least complicated
way possible.

FOLLOWING THE CLUES
PRESENTING A SOLUTION
When you are solving a problem for yourself, you may be able to
obtain a solution without doing any writing other than a few checks
or X's or letters in a table. However, if you wish to present your
solution to others, exhibiting a completed chart is unconvincing. They
will want an explanation of the method of solution. This is best
accomplished by listing with each step or / or X the justification for it.
You can do this by numbering the steps or clues and using these
numbers for reference or as explanation of an entry, as we did in
Figure 1.8. This is a good idea even when you do not intend to
present your solution to anyone else, as it makes it easier for you to
retrace your steps in the event that you find a mistake or want to
check your work. (An illustration of the technique is coming up soon.)
It is also important to be critical of your own work before you
present a solution to someone else. If you are not fully convinced by
your argument, it is not likely that anyone else will be convinced.
SOME STEPS IN PROBLEM SOLVING
We are now ready to list some good techniques to follow in attacking
a problem.
1. As a first step, it is important to understand exactly what information
is given and what is to be found. It is also important to decide what
information is relevant. Sometimes, restating the problem will make
these tasks easier. Often, it is useful either to make a list of the given
information or to use a chart, diagram, or other visual aid.

12 CHAPTER 1
2. After all the given information is listed or charted, add to it all the
information that you can logically deduce from what is given.
3. Continue to add to the list deductively, until either the desired
information is obtained or an impasse is reached—that is, no further
conclusions seem possible but the desired information has not yet
been obtained. (Before you decide that an impasse has been reached,
it is often a good idea to run through all the given clues once again,
bearing in mind what has been found so far. Sometimes this leads
to additional conclusions that had been overlooked.)
4. In the event of an impasse, choose some aspect of the problem for
which only a finite (preferably small) number of nonoverlapping
alternatives are possible. Arbitrarily assume any one of these
alternatives to be true; add it temporarily to the list or chart of facts that
are known; and see what conclusions you can logically draw. If a
contradiction is reached, then conclude that the alternative assumed
is, in fact, not correct and eliminate it as a possible alternative.
Also, add the fact that the alternative is not correct to the list of things
that are known. This fact may now be used in further arguments.
What happens, though, if an assumption does not lead to a
contradiction? It may, for example, lead to an answer. In this case, you cannot
yet conclude that this is the complete solution to the problem. Each
alternative assumption must also be investigated: Every possible case must
be considered. If all these other assumptions lead either to
contradictions or to the same answer found originally, then you can conclude
that this answer is the only solution to the problem. However, if some
alternatives lead to different answers, it follows that either the problem
has more than one solution—and this should be noted—or that the
problem is not well posed. (If the wording of the problem indicates
that there should be a unique solution and if, in fact, there can be
more than one solution, then the problem is not well posed.) If a
problem does have more than one solution, the truly inquisitive mind
would not be satisfied until all solutions are at least categorized, if not
worked out in detail.
Since you must always return eventually to the point at which you
make an assumption—either because the assumption leads to a
contradiction or to consider other alternatives—it is a good idea to enter
that assumption in your visual aid with a pencil, preferably using a
different color from the one you started with. Use the same pencil to
note down all conclusions that are based on the particular assumption.
Then, when you do return to the point of making the assumption,
you will easily be able to eliminate all information based on it, leaving
only what was known before the assumption was made.

FOLLOWING THE CLUES
13
It is possible that after you make an assumption, you still obtain
neither an answer nor a contradiction—you still face an impasse. In
this case, you may have to make a secondary assumption. We will
consider this possibility in more detail in the solution of Problem 1.4.
5. Each solution you obtain should be shown to satisfy completely the
conditions of the original problem. Sometimes an answer appears
when you use part (but not all) of the given information. It is
possible that this " answer " contradicts some of the other conditions.
This is the reason a final check is important, especially if more than
one answer has been obtained.
Now try to apply these techniques to Problem 1.4 (page 3).
The clues are:
1. Each gentleman is the namesake of the profession of one of the
others.
2. The dyer is seated two places to the left of Mr. Hosier.
3. The baker sits two places to Mr. Baker's right.
4. The farmer is seated on Mr. Farmer's left.
5. Mr. Dyer is on the glover's right.
A diagrammatic aid helpful in the solution of this problem is just a
simple picture: the dyer is seated somewhere. Call the chair in which
he is seated chair 1 and number the other chairs in the clockwise
direction. By clue 2, Mr. Hosier is in chair 4 (Figure 1.17).
Solution of
Problem 1.4
(left)
FIGURE 1.17
4 Mr. Hosier
How can we use the other information ? We can chart the numbers
of the seats that could possibly be occupied by each of the people in
clues 3, 4, and 5 (Figure 1.18).

14
CHAPTER 1
By clue 3
By clue 4
Mr. B
1
2
3- -
-f-
5
baker
4
5
- t" <
-^ J
3
Mr. H is in 4
FIGURE 1.18
Mr. Baker
farmer
Mr. Farmer
dyer
FIGURE 1.20
Mr. F
I
2
3
-4^ ~
-5—
farmer
2
3
4
- - 5- 1
—4, <
By clue 5
Mr. D
Mr. H is in 4
dyer is in 1
glover
2 dyer not Mr. D) is in 1
3
4
S Mr. H is in 4
1 dyer is in 1
There are only two possibilities left for Mr. Dyer: seat 2 or seat 3.
Let's investigate each of these alternatives.
Case I Assume he is in seat 3, which means that the glover is in
seat 4. In this case, we can eliminate some more seating possibilities
(Figure 1.19).
In case I:
Mr. B
2
5
FIGURE 1.19
baker
—4— glover is in 4
Mr. F
1
2
3
farmer
2
3
4 1
^ 1
-4- glover is in 4
We have not yet arrived at a solution or a contradiction. However,
at this point, we see that there are only two possibilities for Mr. Baker:
He is in seat 2 or in seat 5. We consider these as subcases, that is,
as alternative secondary assumptions.
Case la Assume that Mr. Baker is in seat 2.
Then, the baker is in seat 5, and Mr. Farmer must be in seat 1 (by
Figure 1.19, since Mr. Baker is assumed to be in seat 2), with the
farmer in seat 2. This gives us Figure 1.20.
By the process of elimination, we can complete the diagram,
obtaining Figure 1.21.
Thus Mr. Farmer is the dyer.
Mr. Dyer
hosier
Mr. Hosier
glover
5 baker
Mr. Baker
farmer
Mr. Farmer
dyer
FIGURE 1.21
Mr. Hosier
glover
Mr. Glover
baker

FOLLOWING THE CLUES
15
Does this complete the solution of the problem? Not yet. Our
conclusion has been based on two assumptions. We must still investigate
the alternatives. We begin with the alternative to our secondary
assumption.
jj
c^^g^fl-
MR.ftlK£KlN2
Case lb Assume that Mr. Baker is in seat 5. (Remember that we are
still working under the primary assumption that Mr. Dyer is in seat 3.)
Then, the baker must be in seat 3, and the farmer must therefore
be in seat 2, with Mr. Farmer in seat 1. This gives the seating
arrangement shown in Figure 1.22.
Again by the process of elimination, we can complete the diagram
to obtain Figure 1.23.
Mr. Dyer
baker
farmer 2
/^ X . Mr. Hosier
f ^ glover
Mr. Farmer
dyer
FIGURE 1.22
1^ ^5 Mr.
Baker
Mr. Glover
farmer
Mr. Farmer
dyer
FIGURE 1.23
Mr. Hosier
glover
Mr. Baker
hosier
Note that although the seating arrangements in Figures 1.21 and
1.23 are not identical, both yield the same answer to the original
question—Mr. Farmer is the dyer.
We are still not yet satisfied. We must return to consider the
alternative to our original assumption.
Case II Assume that Mr. Dyer is in seat 2.
It follows that the glover is in seat 3 (see Figure 1.18). Again we
can use this information to eliminate some of the possibilities from
Figure 1.18. Specifically, Mr. Baker must be in seat 1, with the baker
in seat 4 (see Figure 1.24).

16
CHAPTER 1
Since the dyer is also in seat 1, this seems to give us a different
answer—that Mr. Baker is the dyer. However, we must follow through
to make sure that this situation is consistent with the given
information.
It turns out that it is not, as there is now no seat for Mr. Farmer
(see Figure 1.25).
In case II:
Mr. B
1
c
FIGURE
baker
4
5 A^. Oyer is in 2
-, , - - -»
y ■■ ^ovcr IS in 3
1.24
Mr. F
farmer
-2- Mr. Baker is in 1
-^~ Mr. Dyer is in 2
--4— the baker is in 4
FIGURE 1.25
We have reached a contradiction. Hence, Mr. Dyer cannot be in
seat 2 and must be in seat 3.
This completes the solution of the problem. We have found that
two of the possible seating arrangements (Figures 1.21 and 1.23) are
consistent with the given information. In both, the dyer is Mr. Farmer;
and so that is the answer. (Note that we could also conclude that the
glover is Mr. Hosier, since that is the case in both diagrams; but we
cannot identify the occupations of the other three gentlemen.)
TREE DIAGRAMS
In the solution of Problem 1.4 and the second solution of Problem 1.3,
we reached a point at which we resorted to making assumptions—
that is, considering cases—successively testing the truth of each of
several possible alternatives. However, there are many problems in
which this technique is unnecessary—the solution may be deduced
directly from the given information, without analysis by cases.
On the other hand, in a problem in which we have to assume
something, even after we make that assumption we are not always
able to proceed directly to a solution; we may still have an impasse.
We may then have to consider subcases and make a secondary
assumption, possibly followed by further assumptions. Whenever we
do so, however, we must remember to consider each possible alternative

FOLLOWING THE CLUES
17
to every assumption we make. This procedure, of considering cases
and subcases, is sometimes referred to as case analysis.
To help keep track of these assumptions or cases and their
alternatives, a diagram can be helpful. For example, in Problem 1.4,
after we placed the dyer in seat 1 and Mr. Hosier in seat 4, we saw
in Figure 1.18 that there were two possibilities (2 and 3) for the seat
occupied by Mr. Dyer, three possibilities (1, 2, and 5) for the seat
occupied by Mr. Baker, and three possibilities (1, 2, and 3) for the
seat occupied by Mr. Farmer. We can indicate this diagrammatically,
as in Figure 1.26.
Mr. Dyer's seat Mr. Baker's seat
FIGURE 1.26
This figure is a tree diagram, so called because it branches out as
a tree does. The points at which the branching takes place are called
nodes or branch points. Each line segment connecting two nodes is
called a branch, and a sequence of branches connecting the starting
point of the tree to a terminal point is called a path. Each path
indicates a case to be considered; the whole tree indicates all possible
cases.
The tree in Figure 1.26 indicates all the possible alternatives for
the seats in which Mr. Dyer, Mr. Baker, and Mr. Farmer could be
seated. We read the cases in the tree in the following manner:

18
CHAPTER 1
Case 1 Mr. Dyer is in seat 2, Mr. Baker in seat 1, and Mr. Farmer
in seat 1. (This case cannot actually occur.)
Case 2 Mr. Dyer is in seat 2, Mr. Baker in seat 1, and Mr. Farmer
in seat 2. (This case cannot occur either.)
Case 3 Mr. Dyer is in seat 2, Mr. Baker in seat 1, and Mr. Farmer
in seat 3.
Case 4 Mr. Dyer is in seat 2, Mr. Baker in seat 2 (again an
impossibility), and Mr. Farmer in seat 1.
Etc.
Obviously, many of the branches in the tree do not represent
alternatives that can actually occur. Cases in which two of the three
people occupy the same seat are eliminated immediately. In the
remaining cases, the locations of Mr. Dyer, Mr. Baker, and Mr. Farmer
determine the locations of the glover, the baker, and the farmer
respectively. Figure 1.27 shows this information. In five of the seven
cases, we again find two different people occupying the same seat.
(For each path leading to a contradiction, the people occupying the
same seat are indicated at the right.)
Mr. Dyer glover Mr. Baker baker Mr. Farmer farmer contradiction
^,,^ #3 #4 baker-farmer
^^--#1 #4^"^
^ # 2 # 3 <r ^.^ # 1 #2 glover-baker
^"~^^#3 #4 glover-baker
^,^#2 #3 glover-baker
'#1 #4-^'^
_^#1 #2 y
^#3 #4<^ #2 #5^^
.,---#1 #2 ^
^~^ #2 #3 baker-farmer
FIGURE 1.27
The two cases that remain correspond to the two solutions that we
found when we originally solved the problem. In both, Mr. Farmer
is the dyer.
An alternative tree diagram that more closely describes the procedure
we actually followed in solving Problem 1.4 is shown in Figure 1.28.
Mr. Dyer Mr. Baker
casela—_^2

FOLLOWING THE CLUES 19
THE MULTIPLICATION PRINCIPLE
Many times we are interested in knowing the number of cases that
could possibly arise in a problem, that is, how many paths there would
be in a tree diagram. (Note that this number is the same as the number
of terminal branches.)
In the original tree diagram (Figure 1.26) for Problem 1.4, there are
two branches emanating from the starting points. Each of these splits
into three branches, each of which in turn splits into three others.
How many terminal branches are there altogether?
The answer lies in the Multiplication Principle. The principle
says that, if a tree has m primary branches, each of which splits into n
secondary branches, then there are mn different paths through the tree
(Figure 1.29).
More generally, if a task can be broken up into two steps, the first
of which can be done in m ways and the second of which can be done
in n ways (regardless of how the first step is done), then there are
mn different ways of completing the task.
FIGURE L29
This principle generalizes to tasks that may be broken into more
than two steps. Thus, the tree in Figure 1.26 has three sets of branch
points. At the first or starting point, the tree splits into two branches;
at the second, each branch splits into three; and at the third, each
branch again splits into three. Hence, there are 2 • 3 • 3 = 18 terminal
branches or 18 different paths.
Note that the Multiplication Principle cannot be applied if the number
of secondary branches depends on which primary branch is chosen. For
example, the tree diagram in Figure 1.30 has two primary branches
(aj and ^2)- ^^ ^1 is chosen, then there are only two possible secondary
branches (b, and b2); but if a2 is selected, then there are three secondary
branches (c,, C2, and C3). Thus, the Multiplication Principle does not
apply in this diagram.
The tree diagram approach can also be used to solve Problem 1.5
(page 3).

20
CHAPTER 1
FIGURE 1.30
If you have not yet solved Problem 1.5 (page 3), try it now.
Solution of
Problem 1.5
At the beginning of the game, A has two possible first moves. This
is indicated in Figure 1.31. The upper branch indicates the possibility
that A takes one stick (leaving five); the lower branch shows the case
in which A takes two (leaving four).
number left after
A's move
FIGURE 1.31
We now extend the diagram to show all possible moves by A and B
(Figure 1.32). Subscript notation in the figures shows whose move it is
and which move; for example, A2 means A's second move.
0 (B wins)
0 (B wins)
0 (B wins)
1 0 (A wins)
0 (B wins)
0 (B wins)
0 (Bwins)
0 (A wins)
FIGURE 1.32

FOLLOWING THE CLUES 21
Note that there are 13 different ways in which the game could be
played. Each of these is included as a path from start to finish in the
diagram, and each must be considered in examining the possible
outcome of the game. For example, Figure 1.33 highlights the case in
which A first takes two matchsticks, then B takes one, then A takes one,
and then B takes two.
FIGURE 1.33
Note also that, in the diagram of all the game possibilities (Figure
1.32), some of the branches lead to a win for A while others result in B
winning. Does this mean that we cannot analyze the game—that is,
determine who should be the winner? Not at all. It's true that if both
players play at random, either of them could win; but we have
assumed that the players are intelligent and play to the best of their
abilities. Under this assumption, we can show that one of the two
players (the second one, B, in this case) can be sure of winning.
How can we say this in light of the fact that the tree diagram does
not always lead to the same answer?
The explanation is simple. We must bear in mind that a player has
a choice at each turn. This choice determines the branch that is
followed. Suppose B adopts the following strategy: Always take the
opposite of what A just took. (We will ignore for now the question
of how this strategy was discovered.) Then the game can be represented
by the tree in Figure 1.34. In all cases, B wins. (Note that many
A,
^5
6^^——4
FIGURE 1
B,
i.34
5
-3^
A2
^2-
1
-2-
^1-
B2
-0
u
-0

22
CHAPTER 1
branches of the original tree have been eliminated because B avoids
them.) Thus, we can say that B can always win the game by sticking
to the given strategy—no matter what A does, B has a move that will
eventually lead to a win.
This also shows that A has no winning strategy. No matter what
move A makes, there will always be a branch of the tree leading to a
win for B, namely, the branch chosen by B following the above strategy.
SIMPLIFICATION
So far we have discussed some basic steps for attacking a general
reasoning problem. There are other techniques that are often helpful.
One of the most important is to simplify the problem. This may be
done in many different ways.
One way is to consider special cases of the problem. For example, if we
want to analyze the matchstick game for all possible values of n
(where n is the number of sticks we start off with), we might first try to
solve it for specific values of w, say w = 1, w = 2, w = 3, ..., w = 6, and
then try to generalize for all n.
A second way of simplifying the problem is to reduce it to a
previously solved problem. For example, suppose we want to analyze the
case where the matchstick game starts with seven sticks. Instead of
drawing a tree diagram, we observe that the first player, by taking one
stick, can reduce the game to a six-stick game in which he goes second.
Because we know from before that this game will result in a win for the
second player, we have proven that the first player can force a win in the
seven-stick game.
Now try Problem 1.6 (page 3).
Solution of
Problem 1.6
Another way of simplifying a problem is to break it into parts. Consider
Problem 1.6. Part of the problem is to move car A to the right, and a
second part is to move car B to the left. We simplify the problem by
momentarily shunting car B out of the picture and directing our
attention to the following question: How can the locomotive and car A
switch places in Figure 1.35?
tunnel
FIGURE 1.35

FOLLOWING THE CLUES
23
The solution of this problem is comparatively easy:
1. L pulls A to the right (Figure 1.36).
tunnel
FIGURE 1.36
2. L pushes A to the tunnel (Figure 1.37).
A in tunnel
FIGURE 1.37
3. L goes back to the main track and then moves to the left (Figure
1.38).
A in tunnel
FIGURE 1.38
4. L picks up A at the tunnel and pulls A back to the main track on
the left (Figure 1.39).
tunnel
L A
FIGURE 1.39

24
CHAPTER 1
5. L pushes A to the right and comes back to A's starting position
(Figure 1.40).
tunnel
FIGURE 1.40
Returning to the original problem, we ask how simplifying it can
help us. If we realize that the size of the locomotive plays no role in
the solution of the simplified problem, then the answer is not difficult.
Suppose we first attach L to car B and consider it as one unit LB—as
if it were the locomotive. Then by following the procedure above, we
can move A to the right of LB, not just to the right of L. This
leaves the position shown in Figure 1.41.
tunnel
FIGURE 1.41
If LB now pushes A all the way to the right, we are left with
Figure 1.42.
tunnel
FIGURE 1.42
All that remains for us to do is to switch L and B, which is
essentially the simplified problem revisited. Specifically:
1. L pulls B to the left.
2. L pushes B into the tunnel (Figure 1.43).

FOLLOWING THE CLUES
25
B in tunnel
FIGURE 1.43
3. L returns to the main track and moves to the right (Figure 1.44).
B in tunnel
^Sit^
L A
FIGURE 1.44
4. L picks up B at the tunnel and pulls B back to the main track
at the right (Figure 1.45).
tunnel
i^ap^^rm
B L A
FIGURE 1.45
This completes the problem.
Problem 1.6 is one of a variety of "shunting problems." Among
the exercises in this chapter is an additional problem of this type to
further illustrate the idea of simplification. Several other shunting
problems will be found in Chapter 9.
THE CHAPTER IN RETROSPECT
General reasoning is the basis of mathematics and logic as well as
problem solving in general. This chapter has introduced you to some
principles and techniques used in general reasoning. We have, in an
informal and intuitive way, discussed the basic ideas of making

26
CHAPTER 1
assumptions, looking at alternatives, reaching a contradiction, using
visual aids, and other approaches. Not all the techniques mentioned
here will be applicable to all problems. In fact, for some problems,
none of these will help. However, as you gain more experience with
problem solving, you will develop other techniques.
If you find that you are not making any progress in trying to solve
a problem, it is frequently a good idea to leave the problem for a
while and return to it later, possibly with a fresh attack.
In fact, if you have difficulty with many of the problems, then,
before you attack each one, make a list of all the different approaches
you can think of for that problem. This way, if you should get stuck,
you will already have another point to view from.
Remember that there are sometimes many ways to solve a problem,
and any method that is logically sound is correct.
The remaining chapters of this book are devoted to specific types of
problems and introduce some techniques that will be helpful in solving
them. In particular. Chapter 2 takes a somewhat more formal approach
to the reasoning process by considering some principles of symbolic
logic.
Exercises
1.1 [1 s*
There were five fine ladies from Carruther
Who named their pets after each other.
From the following clues.
Can you carefully choose
The pet which belongs to Sue's mother?
Toni Taylor owns a hog;
Belle Bradkowski owns a frog;
Janet Jackson owns a crow;
The garter snake is owned by Jo;
* @ at the beginning of a problem means you will find
a hint about how to solve it in the Hints and Solutions
seaion (page 340); [s] means the full solution is given
there, [a] means that the answer is given in the section
called Answers to Selected Problems (page 381).
Sue's the name they call the frog;
And **Here Jo, here Jo" brings the hog;
The name by which they call the pony
Is the name of the woman whose pet is Toni;
The final clue, which Til now tell,
Is that Sue's mother's pet is Belle.
1.2 Adam, Robert, Clifton, Stephen, and
Brent are the five starters on the Doylestown
Dribblers basketball team.
Two are left handed and three right handed.
Two are over 6 feet tall and three are under
6 feet.
Adam and Clifton are of the same
handedness, whereas Stephen and Brent use different
hands.
Robert and Brent are of the same height

FOLLOWING THE CLUES
27
range, while Clifton and Stephen are in
different height ranges.
The man who plays center is over 6 feet and
is left handed.
Who is he?
1.3. In the fall, the members of the small
and very exclusive Lawnsand Garden Club
began to plan the following season's garden
arrangements. After several meetings, the five
members of the club (Mr. Isaac Iris, Ms. Rita
Rose, Madame Anastasia Azalea, Dr.
Frederick Forsythia, and Sir Horace Holly)
decided that each should send a plant to one
of the others.
The five plants sent corresponded, in some
order, to the names of these five people.
Each of the five received exactly one plant;
in no case did the receiver, or the sender,
have the same name as the plant.
Ms. Rose sent a holly to Dr. Forsythia.
The recipient of the plant sent by the
doctor sent a rose.
The flower lover with the same name as the
plant sent by Madame Azalea received a
forsythia from the namesake of the plant that
Madame Azalea received.
Who sent what to whom?
1.4. [s] \a\ Allen, Bill, Chuck, Ed, Harry,
Jerry, Mike, Paul, and Sam have formed a
baseball team. The following facts are true:
(a) Allen does not like the catcher, (b) Ed's
sister is engaged to the second baseman, (c)
The center fielder is taller than the right
fielder, (d) Harry and the third baseman live
in the same building, (e) Paul and Chuck
each won $20 from the pitcher at poker.
(f) Ed and the outfielders play cards during
their free time, (g) The pitcher's wife is the
third baseman's sister, (h) All the battery and
infield except Chuck, Harry, and Allen are
shorter than Sam. (i) Paul, Allen, and the
shortstop lost $100 each at the race track.
(j) The second baseman beat Paul, Harry, Bill,
and the catcher at billiards, (k) Sam is in the
process of getting a divorce. (1) The catcher
and the third baseman each have two
legitimate children, (m) Ed, Paul, Jerry, the
right fielder, and the center fielder are
bachelors; the others are married, (n) The
shortstop, the third baseman, and Bill all
attended the fight, (o) Mike is the shortest
player on the team.
Determine the position of each player on the
baseball team.
Note: On a baseball team there are three
outfielders (right fielder, center fielder, and
left fielder), four infielders (first baseman,
second baseman, third baseman, and
shortstop), and the battery (pitcher and catcher).
([27], Jan. 1976, p. 64)
1.5. [h] Six players—Pietrovich, Cavelli, St.
Jacques, Smith, Lord Bottomly, and
Fernandez—are competing in a chess
tournament over a period of five days. Each player
plays each of the others once. Three matches
are played simultaneously during each of the
five days. The first day, Cavelli beat Pietrovich
after 36 moves. The second day, Cavelli was
again victorious when St. Jacques failed to
complete 40 moves within the required time
limit. The third day had the most exciting
match of all when St. Jacques declared that he
would checkmate Lord Bottomly in 8 moves
and succeeded in doing so. On the fourth day,
Pietrovich defeated Smith.
Who played against Fernandez on the fifth
day?
1.6. [h] \a\ Joe Flannery had a week off from
his trucker's job and intended to spend all
nine days of his vacation (Saturday through
the following Sunday) sleeping late. But his
plans were foiled by some of the people who
work in his neighborhood.
On Saturday, his first morning off, Joe was
wakened by the doorbell; it was a salesman
of magazine subscriptions who was "working
his way through college."

28
CHAPTER 1
On Sunday, the barking of the neighbor's
dog as the paper boy tried to deliver the
paper abruptly ended Joe's sleep.
On Monday, he was again wakened by the
persistent salesman but was able to fall asleep
again, only to be disturbed by the garbagemen.
In fact, the salesman, the garbagemen, and
the neighbor's dog that barked at the paper
boy combined to wake Joe at least once each
day of his vacation, with one exception.
The salesman woke him on Wednesday;
the garbagemen on the second Saturday; the
dog on Wednesday and the final Sunday.
No one of the three noisemakers was quiet
for three consecutive days; but yet, no pair of
them made noise on more than one day during
Joe's vacation—for example, the salesman and
the garbagemen both woke Joe on Monday but
did not both wake him up on any other day.
On which day of his vacation was Joe able
to sleep late?
-¥* 1.7. [h] [a\ Five schools competed for the
gold medal in the finals of the Greater
Philadelphia Scholastic Track Meet. They
were Central, Franklin, Lincoln, Mastbaum,
and Olney. The five events in the finals were:
* A star before an exercise means that you may find it
harder to solve than some other problems. Two stars mean
that the problem is a real challenge.
the high jump, shot put, 100-yard dash, pole
vault and one-mile relay. In each event, the
school placing first received five points; the
one placing second, four points; the one
placing third, three points; and so on. Thus,
the one placing last received one point. At
the end of the competition, the points of each
school were totaled; these totals determined
the final ranking.
1. Central won with a total of 24 points.
2. Otto O'Hara of Olney won the high
jump hands down (and feet up), while
Oliver Oats, also of Olney, came in third in
the pole vault.
3. Lincoln had the same number of points
in each of four events.
If each school had only one entry in each
event, and if there were no ties, and if the
schools ended up being ranked in the same
order as the alphabetical order of their names,
then how did Freddy Farkle of Franklin place
in the high jump?
M 1.8. [h] \a\ One afternoon, having nothing
to do. Bob, Carol, Ted, and Alice decided
to have a tiddlywinks tournament. They each
played one game against each of the others.
Each game started with ten tiddlywinks on
the table and ended when all ten had been
flipped into a cup. The players alternated
turns and each scored one point for each
tiddlywink that he or she successfully flipped
into the cup.
Alice was the clear victor, winning all three
of the games she played in and outscoring her
opponents by a total of 22 points.
Carol only won one game, but also outscored
her opponents.
Bob had a better won-lost-tie record than
Ted did; in fact, Bob scored as many points
against Ted as Ted scored in all three of his
games combined.
No two matches ended in the same score.
What was the outcome and the score of each
match?

FOLLOWING THE CLUES
29
1.9. \h\ The Jones family was looking for a
good vacation spot. It had to have a swimming
pool, tennis courts, a golf course, and fine food.
Most importantly, it should not be too
expensive.
Harry's Hideaway, Shangri La, the
Cromwell, and Paradise Lodge were the four
resorts that the Joneses discovered which were
within their price range. Unfortunately, none
of the four had all the other desired
features. However, the Joneses decided to
select the one that met the most of their
requirements.
1. Both Harry's Hideaway and the
Cromwell had an Olympic-size swimming
pool.
2. Both Shangri La and Paradise Lodge
had an 18-hole golf course.
3. The Cromwell and Shangri La had
indoor tennis courts.
4. Paradise Lodge is especially known for
its good food.
5. Of the six possible pairs of
requirements, every pair was met by exactly one
resort. For example, only one resort had
both a swimming pool and tennis courts,
and another had both a swimming pool and
a golf course.
Which resort did the Jones family choose?
^ LIO. \a\ In order to select capable
contestants for their quiz shows, some
producers require prospective contestants to
compete off camera. One show, " Fractured
Funnybones," usually chooses four candidates
and conducts a series of head to head matches
between each pair of them. The candidate who
performs the best is then chosen to appear on
the show.
The game is played as follows: The
contestants are shown a key word, and
whoever first thinks of a joke using that word in
the punch line gets one point. The first
player to get five points wins.
When Manny Morris, Nancy Novokov,
Patti Proctor, and Thomas Twickenham
competed against each other, the results were
as follows:
Patti won all three of her games and her
opponents scored a total of only 2 points
against her.
Nancy scored a total of 10 points and her
opponents also scored 10.
Manny and Thomas both scored 8 points,
but Manny allowed his opponents 15 whereas
Tom's opponents scored only 14. In addition,
Tom scored more points against Patti than
Manny did.
What was the outcome and the score of each
head to head match?
M-¥ Lll. [h] \a\ Good poker players have
four characteristics in common: They are
familiar with the odds associated with card
distribution; they know when it is wise to
bluff; they have poker faces; and they are
lucky.
Angel invited four of his friends to play
poker one evening, around his big circular
table. They were Babs, Cleo, Dot, and Edie.
The following was also true:
Everyone was sitting next to someone who
knew the odds, but four of the five were sitting
next to someone who was not well versed in the
probabilistic aspects of the game.
Four of the people were sitting next to wise
bluffers, but three of them were sitting next
to people who did not know when to bluff.
Four of the players were sitting next to
people with good poker faces, but everyone
was sitting next to someone who could not
keep a straight face.
Exactly three of the people were sitting next
to someone who was noted for good luck.
Each of the players had at least one of the
desirable traits; but only one, the big winner of
the night, had all four.
Babs knows the odds and knows when to
bluff, but does not have a poker face and is
not noted for luck.

30
CHAPTER 1
Edie is not sitting next to anyone who
knows when to blufif.
The person on Dot's right has a poker face.
The person on Cleo's left does not know
when to bluff.
Angel knows the odds, but is not lucky.
Who was the big winner of the evening,
what was the seating arrangement, and which
traits did each of the five players have?
1.12. [a] Lou Lion, Ted Tiger, Eli Elephant,
Gary Gazelle, Peter Panther, and Leopold
Leopard all recently returned from a
photographic safari in Africa. Each "hunter" was
able to photograph only two animals, each of
which happened to be the namesake of one of
the hunter's colleagues.
Each animal was photographed by exactly
two of the men.
Each hunter photographed at least one
member of the cat family; and Eli and Gary
accounted for all four cats between them.
Gary and Peter both filmed a leopard.
The namesakes of the animals that Eli
captured on film both photographed a gazelle.
Ted did not photograph any of the animals
that Peter did.
Gary and Ted had photographs of the same
animal.
Who caught the lion on film?
1.13. [h] \a\ Messrs. Doctor, Lawyer,
Plumber, Gardener, and Butcher met on a cruise
ship and discovered that they all lived within
a 5 mile radius of one another. They also
discovered that the occupation of each is the
namesake of one of the others.
Taking an instant liking to one another, they
exchanged phone numbers, intending to use
each other's professional services when they
returned home. Unfortunately, because they
were all slightly drunk at the time, the phone
numbers got all mixed up. Several weeks later,
when the doctor's pipes began leaking, he tried
to call the plumber but reached Mr. Butcher
instead. And when he tried to call the butcher,
he inadvertently reached Mr. Lawyer. The
number he had written down for the gardener
turned out to belong to Mr. Gardener instead.
But the doctor was not the only one to
make a mistake. Actually, each of the five men
had written down the phone numbers correctly
but had mismatched all of them (except his
own, of course). Furthermore, no two
attributed the same number to the same
person.
The lawyer tried to call Mr. Plumber but
reached Mr. Lawyer.
The butcher tried to call Mr. Plumber but
reached Mr. Butcher.
The butcher also reached Mr. Gardener
when he tried to call the plumber.
Whom did Mr. Butcher reach when he tried
to call Mr. Gardener?
MM 1.14. [s] Three merry logicians were
seated about a table talking, while eating pizza
and drinking beer. Inadvertently, each
smeared his or her face without realizing it.
Suddenly they all looked at each other and
began to laugh. Then one stopped laughing,
for he realized that his own face was smeared.
How did he come to this conclusion?
(Assume that anyone realizing that his or her
own face was smeared would have stopped
laughing.)
MM 1.15. \h\ When the producers of the new

FOLLOWING THE CLUES
31
show, " The Secret Life of Sherlock Holmes,"
were casting for the part of the protagonist,
three young actors applied for the job. They
all gave exemplary readings and the producers
were hard pressed to choose between them.
The director did have a slight preference for
Byron Bentley, who came closest to fitting the
director's image of what the great Holmes
should look like; so it was decided to determine
how Bentley compared to Holmes in the
reasoning department.
The director obtained five identical
handkerchiefs from the wardrobe department and,
in front of the three applicants, wrote
"Sherlock Holmes" on three of the
handkerchiefs and "Professor Moriarty" on the other
two.
He then stood Bentley center stage and
pinned one of the handkerchiefs on Bentley's
back. Next, one of the other applicants was
told to stand behind Bentley and a
handkerchief was pinned on his back. The second
applicant could see the name on Bentley's
handkerchief but could not see the name on
his own. The third applicant was placed
behind the other two, so that he could see the
names on their handkerchiefs but neither they
nor he could see the name on the handkerchief
that was placed on his back.
Bentley, of course, could not see any
handkerchief.
"The first one of you who can deductively
determine the name on his own back will get
the part," the director announced.
After a few minutes of silence, Bentley
correctly asserted that he was Sherlock
Holmes.
How did he know?
"Elementary, my dear reader."
^ ♦ 1.16. [h] [a] On a very busy day, the
four senior partners of the law firm Smith,
Smith, Smith, and Smith had some
sandwiches delivered from the corner
delicatessen. One of the partners had ordered three
salami sandwiches; one had ordered two salami
and one bologna; one had ordered one salami
and two bologna; and the fourth had ordered
three bologna. Each knew that the four orders
were different.
Shortly after they started eating, the owner
of the delicatessen came running in to
apologize for his new waitress, who had
accidentally mixed up the orders so that each
order was placed in the wrong bag.
Ann Smith said, "Oh. I have already eaten
two salami sandwiches, so I know what the
third sandwich in my bag must be."
"In that case," Bob Smith declared, "since
I know what Ann ordered and I have already
eaten one salami sandwich, I know what the
other two sandwiches in my bag must be."
Carla Smith said, "I have not yet opened
my bag but, based on what I have just heard,
I must have received three bologna
sandwiches."
Assuming that the delicatessen owner's
statement implies that each of the four received
the order intended for one of the others, what
did John Smith (who at that moment was out
of the room and had not yet started eating)
order, and what did he receive?
MM 1.17. [s] \a\ The professor was a guest
lecturer at a logic course for senior,
executives of major oil companies. She selected
six male students for this demonstration. The
professor placed fifteen dimes and fifteen
nickels in six tin cups such that each cup
contained the same number of coins but a
different amount of money. She made six labels
showing correctly how much money each cup
held, but attached to each cup an incorrect
label. She explained the situation to the six
students and gave a cup to each. She asked
each man in turn to feel the size of as many
coins as he wanted in his own cup and
announce something true about them. The
only evidence each man had was the size of
the coins he felt, the incorrect label on his

32
CHAPTER 1
own cup, and the statements made by those
who preceded him. The first man said, " I feel
four coins which are not all the same size;
I know that my fifth coin must be a dime."
The second man said, " I feel four coins which
are all the same size; I know that my fifth
coin must be a nickel." The third man said,
" I feel two coins, but I shall tell you nothing
of their size; I know what my other three
coins must be." The fourth man said, " I feel
one coin; I know what the other four coins
must be."
Determine how the remaining two cups
were labeled and what the total value of the
money in those two cups was. ([27], Feb. 1970,
p. 82)
-¥-¥ 1.18. [a] After the senior prom at Merri-
weather High, six friends went to their favorite
greasy spoon restaurant, where they shared a
booth. The group consisted of the senior class
president, the head cheerleader, a player on the
school volleyball team, a player on the
basketball team, the class valedictorian, and the
school principal's only child. Their names
were Bobbie, Frankie, Gerry, Jo Jo, Ronnie,
and Sal, not necessarily in that order.
Each of the six was in love with one of the
others, but no two had crushes on the same
person.
Bobbie was in love with the person sitting
opposite her.
Frankie liked the cheerleader but was sitting
opposite the valedictorian.
Gerry was sitting next to the cheerleader
and was crazy about the class president.
Jo Jo, who was not the valedictorian, was
sitting between the volleyball player and the
class president.
Ronnie disliked the basketball player.
Sal, an orphan, was sitting against the wall
and had a crush on the volleyball player.
The volleyball player sat opposite the
principal's child.
Identify each person's claim to fame.
MM 1.19. [a] The annual mixed doubles
table tennis tourney took place in Paddleford
City in April. The five schools represented
were: East, West, North, South, and Central.
Each team had two members, one male and
one female. The girls were Becky, Emily,
Sylvia, Vicki, and Helen. The five boys were
Irv, Lee, Max, Paul, and Ted.
According to tournament rules, a team is
eliminated after three defeats, and whichever
team survives after the four other teams have
been eliminated is the winner. The tournament
opens on a Monday evening, with three
matches scheduled nightly. If play continues
beyond Thursday, the Friday matches
continue until a winner is determined.
This year's schedule is as follows:
Monday:
1st match, East vs. West.
2nd match. North vs. South.
3rd match. Central vs. the winner of the
first match.
Tuesday:
1st match, the losers of the first two
Monday matches.
2nd match, the winners of Monday's last
two matches.

FOLLOWING THE CLUES
33
3rd match, the winner of the evening's
first match vs. whichever team
hadn't played in either of the
evening's preceding matches.
Wednesday:
1st match, the losers of Tuesday's first
two matches.
2nd match, the winners of Tuesday's last
two matches.
3rd match, the winner of the evening's
first match vs. whichever team
hadn't played either of the
preceding matches.
Thursday:
1st match, the winners of Wednesday's
last twa matches.
2nd match, the losers of Wednesday's first
two matches.
3rd match, the winners of the evening's
two previous matches.
Friday:
1. Emily played in the second match on
both Monday and Tuesday.
2. West won its only match of the tourney
on Monday.
3. Max and his partner participated in five
matches before being eliminated at the
conclusion of the second contest on
Thursday.
4. South played in the first match on both
Tuesday and Wednesday.
5. Helen and her partner won two
matches on Tuesday.
6. Irv and his partner won Wednesday's
second match.
7. Vicki and her partner won Thursday's
first match.
8. Paul and Sylvia were on opposite teams
in Wednesday's third match.
9. Ted and his partner lost Thursday's
third match.
If
10. Central was defeated in Thursday's
first match.
11. The team that handed North its third
defeat played in two more games, but did
not win the championship.
Determine:
(a) The members of each team.
(b) The outcome of Thursday's and
Friday's matches.
(c) This year's championship team. (Al B.
Perlman, [50], Problem 4)
¥ ¥ 1.20. [h] [a] At the end of five events of
the decathlon at an intercollegiate track meet,
Juan Jimenez, Hal Harris, Bill Boone, Michael
Manners, and Tom Twofeathers were in the
first five places (without ties), but not
necessarily in that order. They each wore a
number between 1 and 5 on their jerseys, but
no two wore the same number.
As luck or skill would have it, they also
were the first five finishers (again without ties)
in the sixth event—the high jump.
The five schools they represented were:
Holbrook University, Jupiter College,
Marmaduke University, Mamaraneck
Military Academy, and West Rochedale U. (again,
not necessarily in that order).
Prior to the start of the sixth event, the
competitor from Jupiter was in first place and
the competitor from Holbrook was in fifth.
The competitor wearing # 1 was in third, and
the competitor wearing #5 was in fourth.
Boone's finishing position in the sixth event
was one number lower (one place better) than
his overall standing prior to that event; but
Manners' position in the sixth event was one
number higher than his prior standing. Two-
feathers' finish in the high jump coincided
with Harris' previous standing. Jimenez was
the only competitor whose performance in the
sixth event exactly matched his earlier
performance.
The competitor who was in second after
the first five events was the only competitor

34
CHAPTER 1
whose first initial coincided with the initial of
the first name of his school.
The number worn by the jumper from
Marmaduke was the same as his finishing
position in the high jump; and the number
worn by Harris was one greater than his
standing after five events. Twofeathers wore a
number that was one less than his prior
standing but was equal to the prior standing
of the competitor from Mamaraneck.
Identify each competitor with his school, his
entry number, his standing after the first five
events, and his position of finish in the high
jump. (Adapted from [10], Problem 23)
MM 1.21. [h] [a] There are five houses in a
row (east to west), each of a different color
and inhabited by men of different nationalities,
with different pets and preferences in
beverages and cigarettes.
1. The Englishman lives in the red house.
2. The Spaniard owns the dog.
3. Coffee is drunk in the green house.
4. The Ukrainian drinks tea.
5. The green house is east of the ivory
house and next to it.
6. The Old Gold smoker owns snails.
7. Kools are smoked in the yellow house.
8. Milk is drunk in the middle house.
9. The Norwegian lives in the most
westerly house.
10. The man who smokes Chesterfields
lives in the house next to the man with
the fox.
11. Kools are smoked in the house next
to the house where the horse is kept.
12. The Lucky Strike smoker drinks
orange juice.
13. The Japanese smokes Parliaments.
14. The Norwegian lives next to the blue
house.
Who drinks water? And who owns the
zebra? ([27], Jan. 1971, p. 87)
MM 1.22. [h] [a] Big Data Computer Dating
Service arranged a tour for ten of its members.
Participating in this tour were five young men:
Albert, Barney, Chuck, Danny, and Ernie;
and five young women: Florence, Glenda,
Helen, Inez, and Joan. From the tens of
thousands of clients whose psyches had been
programmed onto the electronic tape, they
were the ones deemed to have the greatest
compatibility potential with all five members
of the opposite sex. One love they all had in
common was theatergoing.
Big Data arranged for them to see five of
the most popular shows on Broadway—
"Kumquats and Kulaks," "The Loquacious
Labradorian," " The Masticating Mahatma,"
" The Narcoleptic Nonconformist," and " The
Omnipotent Ottoman." The theaters in which
these stellar productions were housed were—
not respectively—The Purgatory, The
Quagmire, The Reptilian, The Sarcophagus, and
The Thumbscrew.
The members of the Big Data tour
attended each of the shows with a different
date, and no more than one couple ever
attended the same performance of the same
show. Everybody went to the theater on
Wednesday, Thursday, Friday, and Saturday
evenings as well as to the traditional Saturday
matinee.

FOLLOWING THE CLUES
35
But enough of the prologue! From the
following facts, can you (1) link up each show
with its theater and (2) deduce who saw what
with whom and at which performance?
The facts are:
1. Though Glenda loved both ''Ma-
hatma" and the show at The Thumbscrew,
neither thrilled her quite as much as the one
she saw with Chuck, for this marked her
very first visit to a real Broadway theater.
2. During an intermission, Glenda and
Danny had an argument over which show
had a more exciting last act—"Labra-
dorian" or the one at The Reptilian.
3. As Helen and Danny were leaving the
theater, they agreed that the show they had
just seen wasn't as good as
"Nonconformist" but was somewhat better than the one
at The Quagmire.
4. In a period of some 26 hours, Barney
had theater dates with Florence, Helen, and
Inez (although not necessarily in that order).
5. Chuck's date at "Ottoman" had been
Danny's date at The Reptilian.
6. As Ernie was leaving the theater with
Florence, he remarked that of the five shows
that he had seen that week, he enjoyed
"Kumquats" the most, the one at The
Thumbscrew the least, with the one they
had just seen together falling somewhere in
between.
7. Florence saw "Mahatma" and the
show at The Sarcophagus on the same day.
8. The show at The Purgatory was the
only one on the itinerary that consisted of
one long act without any intermission. Inez
saw it on the same night that Glenda and
Barney were attending "Mahatma."
9. During what may well have been the
most idyllic 26-hour period of her entire
life, Joan saw "Mahatma" and the show at
The Quagmire, and she also had a theater
date with Albert (not necessarily in that
order).
10. Albert's date at The Purgatory had
previously attended "Kumquats" with
Chuck.
11. The last show Glenda saw that week
was "Nonconformist."
12. Neither of the people who attended
a certain performance at The Thumbscrew
together had as yet seen " The Narcoleptic
Nonconformist."
13. During the intermission of
"Ottoman," Chuck and Joan stepped outside for
a breath of air, and they caught a glimpse
of their friend Inez in the intermission
crowd at The Thumbscrew, which was next
door.
14. The night before Inez and Albert had
their date, she had been with Danny, and
he (Albert) saw "Ottoman."
15. Chuck attended The Thumbscrew
with Helen.
16. The Quagmire is across the street
from where "Kumquats" is playing.
17. At the same time that Joan and
Albert had their date, Barney and Florence
saw "Kumquats." (Al B. Perlman, [50],
Problem 2)
1.23. [h] Figure 1.46 shows the same main
tunnel
I[vO~-'-^yD;0~-'-^yI
A B
Io~"-^9-Do~'_yII
C D
FIGURE L46

36
CHAPTER 1
railroad track as shown earlier, but there are
now four railroad cars, A, B, C, and D, and
the locomotive, L, arranged in the order A B
LCD.
(a) Using the side track, rearrange the cars
in this order: C D L A B (Figure 1.47a).
(b) Starting again from the order A B L
C D, rearrange them now in this order: D C
L B A (Figure 1.47b).
tunnel
GI5tE2C
(a^
Jo -pPo -^E^^
D C
tunnel
B A
(b)
FIGURE 1.47

Solve It With Logic
"If it was so, it might be; and, if it were so, it would be; but as it
isn't, it ain't. That's logic." says Tweedledee in Lewis Carroll's
Through the Looking Glass.
What is logic? According to the Britannica World Language
Dictionary, logic is a science—"the science of valid and accurate
thinking." However, it is not only the scientist, the logician, or the
mathematician who is interested in the principles of logic. We all
use some form of logical reasoning regularly. In fact, we have already
had to use it in our analysis of the problems of Chapter 1. Our
approach, however, was informal. At times a little more formality can
be helpful. For example, in elaborately stated problems (such as
Exercise 2.24 in this chapter) the use of a formal or symbolic notation
can greatly simplify the analysis.
In order to attack any problem, it is important to understand the
precise meaning of the problem. This is also true in our everyday lives:
In order for people to communicate efficiently, it is important for
each one to understand precisely what the other means and what
inferences can be drawn. If your mother says, "If I get an income
tax refund, then I'll buy a new television," there should be no
ambiguity about what she means. Does she mean that she will not get
a new television if she does not get a tax refund? If, two months later,
you see a new television in her house, can you conclude that she
received a tax refund?
The rules of logic, which we study in this chapter, will tell us what
she should mean and what conclusions we should be able to draw.
37

38
CHAPTER 2
These rules were first formulated by Aristotle (c. 350 bce). He observed
that many logical arguments have the same form. For example,
consider:
All students are bright people.
All bright people study logic.
Therefore, all students study
logic.
All cows are animals that love
grass.
All animals that love grass moo.
Therefore, all cows moo.
Both have the logical form of syllogism:
All A are B.
All B are C.
Therefore, all A are C.
Aristotle realized that the validity of an argument depends on its
form and not on the specific words that are used. His ideas were
expanded upon by many of the logicians and scholars in science who
followed him. Although some symbols were used by Aristotle and his
followers, it was not until the seventeenth century that the
mathematician Leibnitz attempted to develop a complete algebra of
logic. His efforts went unnoticed for almost two hundred years. Then
new life was injected into the field when Boole, De Morgan, Frege,
and others successfully introduced a symbolic notation that gave great
precision and a new impetus to the subject.
From a recreational point of view, Lewis Carroll helped popularize
logic by presenting many comically worded problems in his various
works. These kinds of puzzles are still popular today and may be
found in the works of H. Phillips [51], G. Summers [61] and [62],
and C. R. Wylie [65], among others.
In this chapter, we study informally some basic aspects of symbolic
logic. We will then see how this symbolism can be helpful in the
solution of certain types of problems. We motivate our discussion
by considering the following sample problems. Try to solve them before
reading the rest of the chapter.
SAMPLE
PROBLEMS
Problem 2.1
On Freshman Day, each new student had the option of being a
Truthteller or a Liar for the day. The Truthtellers had to speak only
the truth; the Liars would speak only lies. I came upon three
freshmen. A, B and C, sitting on a step. I asked A whether he was a
Truthteller or a Liar.
A answered with his back turned, so I could not hear what he said.
"What did he say?" I asked B.

SOLVE IT WITH LOGIC
39
B said, "A says he is a Truthteller."
C said, "B is lying."
Was C a Truthteller or a Liar?
Problem 2.2
Three siblings, Alice, Bob, and Carol, truthfully reported their grades
to their parents as follows:
Alice: If I passed math, then so did Bob.
I passed English if and only if Carol did.
Bob: If I passed math, then so did Alice.
Alice did not pass history.
Carol: Either Alice passed history or I did not pass it.
If Bob did not pass English, then neither did Alice.
If each of the three passed at least one subject and each subject
was passed by at least one of the three, and if Carol did not pass the
same number of subjects as either of her siblings, which subjects did
they each pass?
Problem 2.3
Either Lucretia is forceful or she is creative. If Lucretia is forceful,
then she will be a good executive. It is not possible that Lucretia is
both efficient and creative. If she is not efficient, then either she is
forceful or she will be a good executive.
Can we conclude that Lucretia will be a good executive?
In attacking these and other problems, it is not always necessary
to make formal use of the rules of logic. But it is essential to have a
clear understanding of the statement of the problem and to be able to
determine what conclusions can be deduced from what is stated. It is in
developing these abilities that a study of the rules of symbolic logic
can be helpful.
We begin our study by explaining what is meant by a statement.
STATEMENTS
In the standard theory of logic, a statement or proposition is a
declarative sentence that has one "truth value"—it is either true or
false but not both. If it is true, we say that it has truth value T (true);
if it is false, it has truth value F (false).
For example, the following are statements:
The moon is farther from the earth than it is from the sun.
-^ -- -

40
CHAPTER 2
2 + 3 = 6.
2 + 3 = 5.
George Washington was the first president of the United States.
John Smith will be president of the United States in 1984.
It is now raining in Philadelphia.
Each of the above sentences is a statement, as each has a truth value.
The truth value of each of the first two sentences is F; the next two
have truth value T. Although we cannot (at the time of this writing)
determine the truth value of the fifth sentence, it does have a truth
value—it is either true or false. The truth or falsity of the sixth
sentence will depend on the weather conditions in Philadelphia at the
time that the sentence is uttered. Nevertheless, at any given time, it is
either true or false.
The following sentences are not statements:
Will it rain tomorrow?
Get out of here.
This statement is false.
2 + X = 5.
(see Problem 3 below.)
PRACTICE
PROBLEMS
2A
1. [a] Which of the following are statements?
(a) George Washington was not the first president of the United
States.
(b) Is there life on Mars?
(c) Read this book.
(d) Mr. Dunstan's wife is Inez.
(e) The trash is collected on Mondays and Fridays.
2. Which of the following are statements?
(a) Half past five.
(b) The elevators are not running.
(c) How do I get to City Hall?
(d) The rain in Spain stays mainly in the plain.
(e) I think she's got it.
3. [a] (a) Why is the sentence "2 + x = 5" not a statement?
(b) Why is the sentence "This statement is false" not a statement?
Let us now consider Problem 2.1 (page 38). If you were
unable to solve it before, try it again. Think about what A
actually said.

SOLVE IT WITH LOGIC
41
In order to find out whether or not C is telling the truth, we must
first determine what A actually said. One might expect that what he
said must depend on whether he was a Truthteller or a Liar. But let
us see. Consider each of the possibilities.
If A were a Truthteller, then he would tell the truth and say,
"I am a Truthteller."
On the other hand, if A were a Liar, then he would lie and say,
"I am a Truthteller."
In either case, A must have said, "I am a Truthteller." Hence, B
told the truth and C lied. Thus, C was a Liar.
Solution of
Problem 2.1
T
VARIABLES AND CONNECTIVES
We now return to our discussion of symbolic logic, aided by an
analogy with algebra.
Just as we use the variables x, y, z, . . . , in algebra, to represent
general but unknown numbers, we use the propositional variables
/), ^, r, . . . , to represent general but unknown statements. In algebra,
we may assign a specific value to x, jy, and so on; for example,
"let x= 1." Similarly, p may represent a specific statement. For
example, if we wish p to represent the statement " It is raining," we
write
p: It is raining.
Continuing the analogy, numbers or variables may be combined in
algebra by using the familiar operations ( + , x, -, ^). Similarly,
we may combine statements or propositional variables by using
connectives. The most commonly used connectives are: "and" ( a),
"or" ( v), "if , then " (-), "if and only if" (^), and
"not" (~). For example, if p represents the statement "It is raining"
and q represents " I am carrying an umbrella," then we can form a
new statement " It is raining and I am carrying an umbrella," which
we represent by p Aq.
The five connectives mentioned above enable us to combine
statements to form new ones. Let us examine the logical meaning of these
new, compound statements.
NEGATION
Actually, "not" is not a connective in the same sense as the other
four. Whereas the others are binary connectives—that is, they actually
connect two statements or propositional variables, "not" is unary—it
relates only to one statement or propositional variable, rather than two.
Up is any statement, then "not />" (written symbolically as ^p) is just

42
CHAPTER 2
P I -P
T F
F T
FIGURE 2.1
the Statement "It is not the case that />." The statement ~p is called
the negation of the statement p.
For example, the negation of the statement "It is raining" is the
statement " It is not the case that it is raining," or, more simply, " It
is not raining."
Obviously, if a statement p is true, then its negation ^p is false,
and vice versa. This can be exhibited by means of a visual aid called
a truth table (Figure 2.1).
This table in a sense defines the symbol "--". It tells exactly under
what circumstances --p is true, and under what circumstances it is
false. Observe that the truth value of --p depends only on the truth
value of the statement p and not on the specific statement that p
represents. The table has two lines, as p can have two possible truth
values.
In defining the other four connectives, we deal with two statements,
p and q. Since each can have truth value T or F, there are four
possible cases that must be considered. Figure 2.2 shows these in two
ways.
p <i
\'<
FIGURE 2.2
case 1
case 2
case 3
case 4
P
T
T
F
F
?
T
F
T
F
Note that, by the Multiplication Principle (Chapter 1), there are
eight cases for three statements,p, ^, r; sixteen cases for four statements;
and so on.
"AND"-CONJUNCTION
The meaning of the word "and" is well known. The statement "It is
raining and I am carrying an umbrella," when true, gives us two
pieces of information: one, that it is raining; two, I am carrying an
umbrella.
U S RAININ6 AVJP I AM CARRVIAJG AN MBREULA.

SOLVE IT WITH LOGIC
43
In general, the statement "/> and ^," represented symbolically by
p A ^5 is true whenp and q are both true and only then. This information
may be exhibited visually in a truth table (Figure 2.3).
The statement "/> and ^" is called the conjunction of the statement
pand the statement q-. It is sometimes expressed in English by replacing
"and" with a word such as "but," "however," "moreover,"
"furthermore," or "nevertheless." For example, "Roses are red, but violets
are blue" has exactly the same logical meaning as "Roses are red, and
violets are blue."
p
T
T
F
F
Q
T
F
T
F
pr\q
T
F
F
F
"OR "-DISJUNCTION
The meaning of the word " or" presents more of a problem as it is
used in two different ways in the English language. If Grover says
"Either I lost my wallet or I left it home," he is really asserting that
there are two possibilities:
He lost his wallet.
He left his wallet home.
Grover's intention is to assert that one or the other of the two
possibilities occurred, but not both; that is, the possibility that Grover
left his wallet home and lost it is excluded, and "or" is being used
in the exclusive sense.
On the other hand, the statement "Either the accountant made a
mistake or the teller is a thief" allows three possibilities:
The accountant made a mistake and the teller is honest.
The accountant did not make a mistake and the teller is a thief.
The accountant made a mistake but (and) the teller is a thief
anyway.
In this case, "or" is being used in the inclusive sense; that is, the
speaker is including the possibility that both component parts occurred.
When "or" is used in the inclusive sense, it means that at least one of
the component parts occurs; when it is used in the exclusive sense, it
means that exactly one of the component parts must hold.
This lack of precision in the use of the word "or" can create
problems of communication and understanding. The speaker may
intend the word to have the exclusive meaning, and the listener may
interpret it inclusively; or the other way around.
This ambiguity is not tolerable in the realm of deductive reasoning.
Propositions must be precisely stated in order for us to be able to
draw conclusions from them. For this reason, logicians and
mathematicians have decided to adopt the following convention.
The word ''or" is to be used in the inclusive sense, unless otherwise
indicated. Thus, for any particular statements, /), ^, if we say
"Either p occurred or q occurred," we include the possibility that
both occurred. If we wish to exclude this possibility, we must say
"Either/) occurred or q occurred but not both." We call the statement
FIGURE 2.3

44
CHAPTER 2
p
T
T
F
F
Q
T
F
T
F
^V^
T
T
T
F
"p or ^" used in the inclusive sense the dis|unction of the statement p
and the statement q. We denote it symboHcally by p\/q. Thus, the
truth table of v is as shown in Figure 2.4.
FIGURE 2.4
IT IS RA/A/;>JeoR I AM CARRYING AN UMBRELJJV
CONDITIONAL AND BICONDITIONAL
STATEMENTS
A statement of the form "If p, then ^" is called a conditional
statement. The " if" part is called the antecedent of the statement and
the "then" part is called the consequent. Such a statement is denoted
by p -* ^. Many mistakes in reasoning are made because this type of
statement is often misunderstood.
Imagine that a teacher told his class, " If you pass the final exam,
then you will certainly pass the course." Roberta passed the final exam,
so she can be confident that she will pass the course. But Ruth failed
the final exam. What expectations can she have? The answer is that,
based on the teacher's statement alone, she can have no expectations.
The teacher said what would happen if she passed the final, but not
what would happen if she failed it. If he decides to fail Ruth, there
can be no claim that he acted contrary to his previous statement;
but even if he decides to pass Ruth anyway, he still cannot be accused
of telling a lie.
This is the situation with conditional statements in general. The
statement "If/>, then ^" is easily accepted as true if/> and q both
turn out to be true; and it is clearly false Up turns out to be true and q
turns out to be false. However, if p turns out to be false, then the
statement makes no claims whatsoever about q and hence cannot be said
to be false; for this reason a conditional statement in which the
antecedent turns out to be false is considered to be true regardless of
the truth or falsity of the consequent.
Contrast this situation with the following: The teacher said to Ruth,
"You will pass the course if and only if you pass the final exam."
Now, if Ruth fails the final exam, then she knows that she will fail
the course; and of course, if she passes the final exam, then she has
every right to expect that she will pass the course.

SOLVE IT WITH LOGIC
45
Notice the difference between the conditional "if, then" and the
biconditional "if and only if" statements. The latter, denoted
symbolically by p<-*q, makes a claim regardless of the truth value of p;
whereas the former makes a claim only in the case that the antecedent, p,
is true.
The truth tables for the conditional and the biconditional are shown
in Figures 2.5 and 2.6.
p
T
T
F
F
9
T
F
T
F
p-^q
T
F
T
T
FIGURE 2.5
p I Q \P*-*q
T T I T
T F F
F T F
F F T
JS RAINlti6 IF AUDCNLVIF I fMCAf<R^]KaM UM&REuA.
FIGURE 2.6
There are many different ways in which a conditional statement can
appear in English. For example, each of the following statements has
the same logical meaning as "if/), then q'':
q, if/)
p only if q
/) is a sufficient condition for q
^ is a necessary condition for p
q, whenever p
q, provided that p
not p, unless q*
q, unless not p
* It will be shown (see observation 9 on page 48) that an "If , then "
statement can also be translated as an "or" statement. Thus "not p, imless ^" can be
translated as "q or not p." For this reason, the word "unless," like the word "or," is
sometimes used in more than one sense. For example, the statement " I won't go unless
you go" might be interpreted in the biconditional sense " I will go if and only if you go."
However, in this text the word "unless" is to be used in the conditional sense: "If you
don't go, then I won't," which, as we'll see below (page 54), is equivalent to " If I go,
then you go."

46
CHAPTER 2
Similarly, the biconditional statement "p if and only if ^" sometimes
appears in the form "/> is a necessary and sufficient condition for ^."
PRACTICE
PROBLEMS
2B
. [a] Suppose that
/>: Jarvis received a B grade on the final examination
q\ Jarvis passed the course
are both true statements. Find the truth values of each of the
following statements:
(a) If Jarvis received a B on the final, then he passed the course.
(b) If Jarvis received an F on the final, then he failed the course.
(c) If Jarvis received an F on the final, then he passed the course.
(d) If Jarvis passed the course, then he received an A on the final.
(e) If Jarvis failed the course, then he received an A on the final.
Suppose that
/>: Chico earns $4.50 an hour
q: Chico makes more money per week than Terry does
are both true statements. Which of the following are true?
(a) If Chico earns $3.00 an hour, then he does not make more
money per week than Terry does.
(b) If Chico earns $4.50 an hour, then Terry makes more money
per week than Chico does.
(c) If Chico makes more money per week than Terry does, then
Chico earns $4.50 an hour.
(d) If Chico makes more money per week than Terry does, then
Chico earns $5.00 an hour.
(e) If Terry makes more money per week than Chico does, then
Chico earns $5.00 an hour.
3. [a] Write each of the following statements symbolically and then
express each in English in an "if , then " form:
(a) The paraffin test will be positive (r) provided that you recently
fired a gun (/).
(b) I will go to the beach tomorrow C?), unless it rains (r).
(c) Hearing you say so {h) is sufficient for me to believe it {b).
(d) rU leave (/)> if you ask me nicely (a).
(e) Tex cries (c) whenever he sees a cowboy movie (5).
4. Write each of the following statements symbolically and then express
each in English in an "if , then " form:
(a) You must eat {e) if you want to grow {g).

SOLVE IT WITH LOGIC 47
(b) It is necessary for you to acquire a driver's license (a) in order
for you to be permitted to drive a car {d).
(c) A rectangle is a square (r) only if all four sides are the same
length (5).
(d) Being a multiple of four (m) is a sufficient condition for a
number to be even {e).
5. [a] If/) and q are the statements
p\ Alice passed math
q: Bob passed math,
translate each of the following into symbols and indicate in which
cases each is true:
(a) If Alice passed math, then so did Bob.
(b) If Bob passed math, then Alice did.
(c) Alice and Bob both passed math.
6. If r and s are the statements
r: Alice passed history
5: Carol passed history,
translate each of the following into symbols and indicate in which
cases each is true:
(a) Either Alice passed history or Carol did not pass it.
(b) Alice did not pass history.
(c) Alice did not pass history unless Carol passed it.
7. [a] If/) and q are the statements
p: Jack is a good golfer
q: Jack is not a good tennis player,
translate each of the following into English:
(2i)p^q (b)pAq (c) ^q (d) -/) (c)q^p.
8. If r and s are the statements
r: Macbeth is the Thane of Cawdor
5: Ruth is the Sultan of Swat,
translate each of the following into English:
(a)r-"5 (b) 5-► r (c) ~r (d)rv5 (e) r<-^5.
DRAWING CONCLUSIONS
We are now ready to discuss ways in which we can draw conclusions
from statements of the type above. These conclusions or observations
follow directly from the truth tables defining the given statements.

48
CHAPTER 2
From the definition of ~ we make the following observations:
Observation 1. Either/) is true or ^p is true (not both).
Observation 2. It is not possible for both p and ^p to be true
simultaneously. (Thus, if an assumption leads to a situation where p
and ^p are both found to be true, then we say that the assumption
has led to a contradiction and the assumption must be false.)
The definition of a leads to
Observation 3. If/> a ^ is true, then p must be true and q must be true.
Observation 4. If/> a ^ is false, then at least one of/), q is false. Hence,
in particular, \i p /\q is false and p is true, then q must be false; and,
similarly, \i p /\q is false and q is true, then p must be false.
The definition of v leads to
Observation 5. If/> v ^ is false, then p must be false and so must q.
Observation 6. If/> v ^ is true, then at least one of/), q is true. Hence,
in particular, if pv q is true and /> is false, then q must be true; and,
similarly, if pv q is true and q is false, then p must be true. (Note,
however, that if pv q is true and if p is true, we cannot conclude
that q is true or that it is false.)
The definition of -► leads to
Observation 7. If p ^ q is true and p is true, then q must be true.
(Note, however, if p ^ q is true and q is true, p could be true or false.)
Observation 8. If p ^ q is true and q is false, then p must be false.
Observation 9. If p -^ q is true, then either p is false or q is true, or
both. (Note that we cannot in this case conclude explicitly that/) is false
and q is true.)
Observation 10. Up -^ qis false, thenp must be true and q must be false.
Finally, from the definition of <-> we can observe that
Observation 11. If/) <-> ^ is true, p and q must have the same truth value.
Observation 12. If/)<-+^ is false, then p and q have opposite truth
values.
PRACTICE
PROBLEMS
1. \a\ Giventhat"If Archie was early, then no one was home "is a true
statement, what can we conclude if we find out that
(a) Archie was early.
(b) Someone was home.

SOLVE IT WITH LOGIC
49
Given that "Bonnie is tired and Carrie is sleeping" is a false
statement, what can we conclude if we find out that
(a) Bonnie is tired.
(b) Carrie is sleeping.
(c) Bonnie is tired if and only if Carrie is sleeping.
COMPOUND STATEMENTS
As we mentioned earlier, statements and connectives can be used as
building blocks to form new statements. For example, from the simple
statements
e\ Lucretia is efficient
/: She is forceful
g: She will be a good executive,
we can form the compound statement, "If Lucretia is not efficient,
then either she is forceful or she will be a good executive."
This statement may be represented symbolically as
Note the importance of the parentheses to indicate exactly which
statements are connected by each connective.
The types of conclusions that we were able to draw in the preceding
section can be extended to more complicated statements such as the one
above. For example.
If the above statement is known to be true, and if we find out
that Lucretia is not efficient, then we can conclude that either she is
forceful or else she will be a good executive.
If the given statement is known to be true, and we find out that
Lucretia is not forceful and that she will not be a good
executive, then we can conclude that she must be efficient.
. [a] If/), ^, r, and s are the statements
/>: George Washington slept here
q: George Washington is the father of our cbuntry
r: Paul Revere woke George Washington
s\ The Declaration of Independence was signed on July 4, 1776,
translate each of the following into symbols:
(a) If George Washington is the father of our country, then he
didn't sleep here.
(b) If the Declaration of Independence was signed on July 4, 1776,
then either George Washington did not sleep here or else Paul
Revere woke him.
PRACTICE
PROBLEMS
2.D

50 CHAPTER 2
(c) Paul Revere did not wake George Washington unless George
Washington slept here.
(d) If Paul Revere woke George Washington if and only if George
Washington slept here, then the Declaration of Independence
was signed on July 4, 1776 and George Washington is not the
father of our country.
2. If/), q^ r, and s are the statements
p: Roses are red
q: Violets are blue
r: Sugar is sweet
s\ You are sweet,
translate each of the following into symbols:
(a) If roses are red and violets are blue, then sugar is sweet and so
are you.
(b) You are sweet and sugar is too, unless roses are not red or violets
are not blue.
(c) Roses are red only if violets are blue, but if sugar is sweet then
you are not sweet too.
3. [a] If/), ^, r, and s are the statements
p: The poet is dreaming
q: The poet is jesting
r: Titania is the queen of the fairies
5: Oberon is the king of the fairies,
translate each of the following into English:
(a)/)-(^r) (b)(^/))v(^r) (c)^(pAr)
(d)[^(pv^)]-(5-r)
4. If/), ^, and r are the statements
p: Jack and Jill went up the hill
q\ Little Bo Peep has lost her sheep
r: Little Tommy Tittlemouse lived in a little house,
translate each of the following into English:
(a)(^/))^(-^) (b) ^(pv^) (c)(^-/))-r
5. [a] Given that "If either Archie is early or Janice is absent, then
no one is home" is a true statement, what can we conclude if we
find out that
(a) Archie is early.
(b) Someone is home.
(c) Either Archie is early or no one is home.
6. Given that " If today is Thursday and the sun is shining, then
either the robins are singing or it is not summer" is a true statement.

SOLVE IT WITH LOGIC
51
what can we conclude if we find out that
(a) It is summer and the robins are not singing.
(b) Today is Thursday and the robins are singing.
(c) Today is Thursday and the robins are not singing.
[a] If it's true that three krimmls are worth one glunk and that if
Xcag Zemph is the ruler of Mars then the Martian canals are empty,
then it's also true that I am not a Martian. Given that the above
statement is true, what can we conclude if we find out that
(a) Xcag Zemph is the ruler of Mars and I am a Martian.
(b) The Martian canals are empty and I am a Martian.
(c) Three krimmls are worth one glunk and the Martian canals are
empty.
(d) Three krimmls are worth one glunk and Xcag Zemph is the
ruler of Mars.
(e) Three krimmls are worth one glunk and I am not a Martian.
We are ready to discuss the solution of Problem 2.2
(page 39). If you were unable to solve it before, try it again
before reading on.
For reference purposes, we label the statements as follows:
1. Alice: If I passed math, then so did Bob.
2. Alice: I passed English if and only if Carol did.
3. Bob: If I passed math, then so did Alice.
4. Bob: Alice did not pass history.
5. Carol: Either Alice passed history or I did not pass it.
6. Carol: If Bob did not pass English, then neither did Alice.
7. Each of the three siblings passed at least one subject.
8. Each subject was passed by at least one of the three siblings.
9. Carol did not pass the same number of subjects as either of her
siblings.
What conclusions can we draw from these statements about each of
the three subjects?
Statements 1 and 3 deal with math: Since statement 1 is true, it
follows, by observation 7 (page 48), that if "Alice passed math" is
true, then "Bob passed math" is also true. Similarly, by statement 3
and observation 8, if "Alice passed math" is false, then "Bob passed
Solution of
Problem 2.2

52
CHAPTER 2
math" is also false. Therefore, since statements 1 and 3 must be true
simultaneously, there are two possibilities:
Alice and Bob both passed math.
Neither Alice nor Bob passed math.
This information can be recorded in a chart (Figure 2.7).
A
B
C
M
y
y
A
B
c
M
X
X
FIGURE 2.7
Now consider statements 2 and 6, which pertain to English. By
statement 2, together with observation 11, exactly two cases are
possible:
Alice and Carol both passed English.
Neither Alice nor Carol passed English.
These conclusions may be combined with the previous case to give
four possibilities (Figure 2.8):
A
B
C
M
y
y
E
y
y
II
A
B
C
M
y
y
E
X
X
A
B
C
M
X
X
E
y
y
IV
A
B
C
M
X
X
E
X
X
FIGURE 2.8
Statement 6 together with observation 9, implies that either:
Bob passed English (and Alice did or did not pass it)
or
Alice did not pass English (and Bob did not either).
Of these, only the case that both Bob and Alice passed English is
consistent with charts I and III of Figure 2.8 (since, in these charts,
Alice passed English). Statement 6 does not lead to any conclusion
about charts II and IV. However, statement 8 does: Since at least
one of the three siblings passed English, Bob must pass English
(Figure 2.9).
I
A
B
C
M
y
y
E
y
y
y
A
B
C
M
y
y
E
X
y
X
III
A
B
C
M
X
X
E
y
y
y
IV
A
B
C
M
X
X
E
X
y
X
FIGURE 2.9

SOLVE IT WITH LOGIC
53
Now, for history: By statement 4,
Alice did not pass history.
Since statement 5 is true, it follows from observation 6 that
Carol did not pass history.
This information, together with statement 8, implies that
Bob passed history.
We now have Figure 2.10.
A
B
C
M
y
y
E
y
y
y
H
X
y
X
A
B
C
M
y
y
E
X
y
X
H
X
y
X
III
A
B
C
M
X
X
E
y
y
y
H
X
y
X
IV
A
B
C
M
X
X
E
X
y
X
H
X
y
X
FIGURE 2.10
In this figure, chart IV violates statement 7 and so cannot hold.
By applying statements 7 and 8 to charts II and III, we obtain
Figure 2.11.
A
B
C
M
y
y
E
y
y
y
H
X
y
X
II
A
B
C
M
y
y
y
E
X
y
X
H
X
y
X
III
A
B
C
M
X
X
y
E
y
y
y
H
X
y
X
FIGURE 2.11
In charts II and III of this new figure, statement 9 is violated.
Only chart I remains, and, by statement 9, Carol must pass only one
subject (Figure 2.12).
A
B
C
M
y
y
X
E
y
y
y
H
X
y
X
FIGURE 2.12
Thus, Bob passed all three subjects, Alice passed math and English,
and Carol passed only English. You should now check that this
solution is consistent with all nine statements.
LOGICAL IMPLICATION AND EQUIVALENCE
So far we have discussed the logical meaning of certain terms and
how we can draw logical conclusions from statements involving them.

54
CHAPTER 2
If we can logically deduce »S from knowing R, then it must follow that 5
is true whenever R is true; that is, all conceivable circumstances that
would make R true must also make 5 true. We say that R logically
implies S, and write this symbolically as R=> S.
For example, in our previous discussion of drawing conclusions, we
saw that if p -^ q is true and if p is true, then we can conclude that
q is true; that is,
[(p^q)Ap]^q.
Similarly we can express each of our other observations above as a
logical implication. For example, observation 6 (page 48) becomes
[(P V ^) A ( -p)] => q.
Sometimes, we have two statements R and »S such that not only
does R logically imply »S but also »S logically implies R\ that is,
whenever R is true, then »S is true and whenever »S is true, then R is
true. In this case we say that R and «S are logically equivalent
and we express this symbolically as R<=> S,
For example, the statements p ^q and (^q)-'(^p) are logically
equivalent. One way to see this is to consider truth tables (Figure 2.13).
Observe that p-^ q and (~^)-^(~/>) are true under exactly the
same circumstances.
P\q_\p^
T T T
T F F
FT T
F F T
(-q)^(-p)
T
F
T
T
FIGURE 2.13
We can also see the logical equivalence of p -^ q and (^q) -^(~/>)
without actually drawing truth tables: The statement p ^ q is true
except in the case that p is true and q is false; the statement
(^q) -^ (^p) is true except in the case that (^q) is true and (~p) is
false—that is, p is true and q is false. Thus these statements are true
in exactly the same circumstances, and hence are logically equivalent.
The statement (^q)^(^p) is called the contrapositive of the
statement p -^ q. Since they are logically equivalent, these two
statements are identical in meaning and may be used interchangeably.
PRACTICE
PROBLEMS
2.E
ji/^
1. \a\ State the contrapositive of each of the following statements:
(a) If Jonas has at least two nickels in his wallet, then he has at
least ten cents.
(b) If the door is locked, no one can enter.

SOLVE IT WITH LOGIC
55
(c) If I go to the concert, Fll hear some music.
(d) If the three angles of a triangle are equal, then the triangle is
equilateral.
2. State the contrapositive of each of the following statements:
(a) If you sit in the sun, then you get a tan.
(b) If Juanita reads the newspaper, she will know the news of the
day.
(c) If the measure of an angle is 30^, then the angle is acute.
(d) If my wife is not home, then no one answers the phone.
3. [a] Which of the following statements are equivalent to each other?
(a) If Sondra attended the lecture, she fell asleep.
(b) If Sondra didn't attend the lecture, she didn't fall asleep.
(c) If Sondra fell asleep, then she attended the lecture.
(d) If Sondra didn't attend the lecture, then she fell asleep.
(e) If Sondra didn't fall asleep, then she didn't attend the lecture.
4. Which of the following statements are equivalent to each other?
(a) If it will rain, then the roof will leak.
(b) If it won't rain, then the roof won't leak.
(c) If the roof won't leak, then it won't rain.
(d) If the roof will leak, then it will rain.
(e) If it won't rain, then the roof will leak.
ARGUMENTS AND VALIDITY
In most situations that require deductive reasoning, we are given
certain clues or information and are required to determine what
conclusions can logically be drawn from this information: That is, we are
given certain statements P^, P2 > • • • > ^/j ^^^ ^re required to find another
statement, C, which follows from the given statements.
In general, a set of statements P,, P2 > • • > ^/j> ^> o^c of which (C)
is alleged to follow from the others, is called an argument. The
statement, C, which is alleged to follow from the others is called the
conclusion of the argument; the other statements (P^, P2, ..., Pk)
are called the premises. We sometimes indicate an argument as follows
p.
Pk
means "therefore.")

56 CHAPTER 2
For example.
If Lucretia is forceful, then she will be a good executive
Lucretia is forceful
.'. Lucretia will be a good executive.
is an argument. Its premises are " If Lucretia is forceful, then she will
be a good executive" and "Lucretia is forceful"; and its conclusion is
"Lucretia will be a good executive."
If the conclusion of an argument does logically follow from the
premises, then we say that the argument is valid. A valid argument
is one in which the conclusion must be true whenever all the premises
are true. This may be expressed in terms of logical implication:
(P,AP2A'"APk)=>C.
In general, to show that an argument is valid, we must show that
the conclusion is true in all cases in which the premises are all true.
Thus, we assume that all premises are true and we must show that it
then follows that the conclusion is true.
To show that an argument is not valid, it is enough to find one
case in which all the premises are true but the conclusion is false.
For example, the argument
" If Lucretia is forceful, then she will be a good executive.
Lucretia is forceful. Therefore, she will be a good executive."
is valid. To see this, let p and q represent the statements
p: Lucretia is forceful
q: She will be a good executive.
Then the argument is of the form
p-^q
P
As we have already seen that [(p -^ ^) a/)] => ^, this argument is valid.
On the other hand, the argument
" If Lucretia is forceful, then she will be a good executive.
Lucretia will be a good executive. Therefore, Lucretia is forceful."
is not valid. It has the form
p-^q
Q
But here it is possible for both premises to be true while the
conclusion is false. Specifically, if/) is false (Lucretia is not forceful)
and q is true (she will be a good executive), then both premises are
true but the conclusion is false. Hence the argument is not valid.

SOLVE IT WITH LOGIC
57
Note that the vaHdity of an argument depends only on the form
of the argument and not on the specific statements involved. Thus,
for example, any argument that can be represented in the form
P
is valid, and arguments of the form
P-
Q
are not valid.
As an example of a form of a valid argument involving three variables,
consider the following:
q-^r
'• p-^r
This form of argument is important because it not only appears often
in everyday reasoning but also is used in most mathematical proofs.
We leave it as an exercise for you to show that arguments of this
form are valid.
[a] What are the premises and what is the conclusion of each of the
following arguments? Represent each argument symbolically.
(a) Either Rachelle is brilliant or she has a wonderful personality.
If she has a wonderful personality, then she has an active social
life. Therefore, either Rachelle is brilliant or she has an active
social life.
(b) If Bessie is a cow, then she moos. Bessie is not a cow. Therefore,
Bessie doesn't moo.
(c) Either Lucretia is forceful or she is creative. She is forceful.
Therefore, she is not creative.
(d) If Lee Wong is a student, then he is bright. If Lee Wong is
bright, then he studies logic. Therefore, if Lee Wong is a student,
then he studies logic.
What are the premises and what is the conclusion of each of the
following arguments? Represent each argument symbolically.
(a) Either Lucretia is forceful or she is creative. She is not forceful.
Therefore she is creative.
(b) If Lucretia is not efficient, then either she is forceful or she will
PRACTICE
PROBLEMS
2.F

58 CHAPTER 2
be a good executive. Lucretia is not forceful. Lucretia will not be
a good executive. Therefore, Lucretia is efficient.
(c) If Bessie is a cow, then she moos. Bessie moos. Therefore,
Bessie is a cow.
(d) If Bessie is a cow, then she moos. Bessie doesn't moo. Therefore,
Bessie is not a cow.
(e) If Bessie is a cow, then she moos. Bessie is not a cow. Therefore,
she doesn't moo.
3. [a] Determine whether each of the following is the form of a valid
argument:
(a) pv^ (b) p-^q (c) pvq (d) p-^q
-/> q-^r p -/>
.'. q .\ P^r :. ^q /. q
4. Determine whether each of the following is the form of a valid
argument:
(a) p-^q (b) -(pA^) (c) p-^q {d) p-^q
^q P q-^p q-^P
.*. p .'. ^q :. pAq :. p*-*q
5. \a\ For each of the arguments in Problem 1, determine whether or
not it is valid.
6. For each of the arguments in Problem 2, determine whether or not
it is valid.
7. [a] For each of the indicated conclusions, determine whether the
argument is valid:
If Alice passed math, then so did Bob
If Bob passed math, then so did Alice
Therefore
(a) Alice and Bob both passed math.
(b) Alice passed math if and only if Bob passed it.
8. For each of the indicated conclusions, determine whether the
argument is valid:
Alice passed English if and only if Carol did
If Bob did not pass English, then neither did Alice
Therefore
(a) If Bob passed English, then Carol did not.
(b) If Bob passed English, then Carol did.
(c) If Bob did not pass English, then neither did Carol.

SOLVE IT WITH LOGIC
59
9. Show that arguments of the form
are valid. (Hint: Under what circumstances will the conclusion be
false? Is it possible for both premises to be true in this case?)
We are now ready to solve Problem 2.3 (page 39). If you
haven't already solved it, try it again now.
In this problem we are given certain information and asked whether
or not a particular conclusion follows from this information; that is,
we are being asked if a particular argument is valid.
We present two methods of attacking this problem: on the left, an
informal, verbal approach; on the right, a symbolic approach. In the
latter approach, we begin by representing the simple statements of the
problem by symbols.
Verbal approach
p.
The premises of the argument
are:
Either Lucretia is forceful or
she is creative.
If she is forceful, then she
will be a good executive.
Lucretia is not both efficient
and creative.
If she is not efficient, then
either she is forceful or she
will be a good executive,
desired conclusion is:
C: Lucretia will be a good
executive
We may assume that all the
premises are true. Therefore, by
Pj, there are two cases to
consider:
P.
The
Symbolic approach
/: Lucretia is forceful
g: Lucretia will be a good
executive
e: Lucretia is efficient
c. Lucretia is creative
The premises may be expressed
as:
P,:/vc
P,: ^{e/\c)
and the conclusion to be tested:
C:g
We may assume that all the
premises are true. Therefore, by
P,, there are two cases to
consider:
Solution of
Problem 2.3

60
CHAPTER 2
Case 1. Lucretia is forceful.
Case 2. Lucretia is not forceful.
Suppose Lucretia is forceful.
Then, by the second premise,
she will be a good executive (and
so the desired conclusion
follows).
If Lucretia is not forceful,
then, from the first premise, she
must be creative.
Since she cannot be both
efficient and creative (by P3), she
cannot be efficient.
Therefore, by P4, either she is
forceful or she will be a good
executive.
Since we have assumed that
Lucretia is not forceful, we can
conclude that she will be a good
executive.
Case 1./ is true.
Case 2. / is false.
In case 1, it follows from P2
(by observation 7, page 48) that
g must be true, and so the
desired conclusion follows.
In case 2, it follows from P^
(by observation 6) that c must be
true.
By P3 and observation 4, e
must be false.
But, then, by P4 and
observation l^jyg is true.
But, in case 2, / is false and
so, by observation 6, g is true.
In both cases the conclusion is true and so the argument is valid.
In this problem we did not seem to gain very much by representing
the problem symbolically. However, for problems in which the wording
is very involved, the use of letters to represent statements can help
us eliminate the complicated language and get down to the basics of
the problem. It is for this reason that we have introduced the
symbolic approach above.
THE CHAPTER IN RETROSPECT
This chapter has been a brief and informal introduction to symbolic
logic. You have been introduced to the logical meanings of the
connectives and some of the basic rules of inference, so that you will
be able to determine whether or not particular conclusions follow from
a given set of statements. Obviously, this knowledge is important to
the reasoning process.
In applying the principles of symbolic logic, you seldom have to
point them out. As long as you are satisfied that each conclusion
you draw logically follows from what you knew previously (and that
this could be justified formally if it were necessary to do so), you

SOLVE IT WITH LOGIC
61
can usually suppress formalism. In attempting the exercises at the
end of the chapter, just be sure that you are completely convinced by
each step of your arguments. You may find that representing statements
symbolically may prove helpful; if this is the case, don't hesitate to
do so. And, of course, to present your solutions to others, you must
explain your reasoning, and symbols are often helpful in doing so.
Notice that no attempt has been made in this chapter to list a set
of steps that must be followed to reason through every problem with
which you may be faced. Every problem must be attacked in its own
way.
Nevertheless, to attack any problem no matter what the setting (in
business, social science, economics, etc.), you must clearly understand
the problem and must be able to recognize when one statement follows
from another. It is in carrying out these tasks that this chapter may
prove helpful.
Exercises KLI
Truthtellers and Liars
2.1. One summer solstice, while searching
for the Southern Shortcut, Silas Seeker, that
superman of seafarers, was shipwrecked
during a sudden severe squall. Auspiciously,
Silas sighted a small sandy isle in the distance
and succeeded in swimming ashore.
Exhausted, he slipped into a sound sleep.
While he slept, Silas dreamt that he was
discovered by two trolls, both of whom,
fortunately, were able to converse in English.
The two were identical in all aspects of their
appearance; however, one belonged to a clan
whose members always tell the truth, the other
to a clan whose members always lie.
When Silas awoke, he discovered two trolls
standing over him, exactly as in his dream.
"Where am I?" he inquired.
"The Isle of Hamlock," replied the first
troll.
"The Isle of Grindle," replied the second.
"And what are your names?" asked Silas.

62
CHAPTER 2
"I am Glog and he is Glum," responded
the first.
" No, I am Glog and he is Glum," answered
the second.
Just then a third troll appeared. Hoping to
shed some light on the matter, Silas pointed
to the two and asked, " Which of them can I
believe?"
" He and I belong to the same clan," replied
the third troll, pointing to the first.
"That's true; they do belong to the same
clan," said the second troll.
Assuming that Silas' dream was accurate,
who is telling the truth and what was the
name of the island?
2.2. [a] Silas lived on the island for six years
(see Exercise 2.1), but during all that time he
never learned how to visually tell the difference
between members of the two clans.
One day Silas met two trolls. The first
claimed that they belonged to different clans,
whereas the second proclaimed that the first
was a liar.
To which clan did each of the two belong?
2.3. [h] During his sojourn on the island, Silas
gave names to the two clans (see Exercises 2.1
and 2.2). The truthtellers he called Truthfuls
and the liars he called Liars.
One day, Silas met three trolls. He inquired
to which clans they belonged.
"All of us are Liars," said the first.
"No, only two of us are Liars," amended
the second.
" That's not true either," corrected the
third. "Only one of us is a Liar."
To which clan does each of the three belong?
M 2.4. [s] [a] Shortly after Silas completed
building a raft to take him off the island (see
Exercises 2.1 to 2.3), five trolls arrived with
food and water for his voyage. As usual, Silas
inquired about their clans.
"Three of us are Truthfuls," said the first.
"That's right, three of us are Truthfuls,"
the second agreed.
" No, only two of us are Truthfuls," said the
third.
"The first three are all lying," said the
fourth.
The answer of the fifth man left no doubt in
Silas' mind as to the clan affiliation of each
troll. To which clan did they each belong?
2.5. [a] One hundred and fifty years after
Silas Seeker left the island (see Exercises 2.1
to 2.4), his great-great-granddaughter, Heidi
Seeker, found Silas' diary and decided to
revisit the site of her ancestor's salvation.
When she arrived on the island, she was
surprised to discover that there were now three
clans instead of two. During the century and a
half that had elapsed, some intermarriage
between the two groups had occurred and the
progeny of these mixed marriages comprised a
third group of people who always alternated
between telling the truth and lying. Thus,
when a member of this third group (whom
Heidi named Alternators) makes a series of
statements, the first statement might be
truthful or a lie, but the veracity of each subsequent
statement would alternate. In other words, if
the first statement were truthful, then the
second would be a lie, the third would be true,
and so on.
When Heidi first arrived on the island, she
encountered a group of three trolls. One was
very tall, one was very short, and the third was

SOLVE IT WITH LOGIC
63
of medium height. Heidi introduced herself.
" Welcome," said the shortest of the three.
" I am Gladden. He (indicating the tallest of
the three) is Lowax. And he (indicating the
remaining troll) is Grout."
"He is, as he says, Gladden. But I am
Kaut," corrected the troll of medium height.
"And the tall troll is Tildeau."
The tall troll objected: "I am Lowax. But
the short one is Waldar. And he (pointing to
the troll of medium height) is Gaut."
If at least one of them was a Truthful,
what were the names of the three trolls?
2.6. [h] Later that day, Heidi (see Exercise
2.5) met another group of three trolls—one
from each clan. Heidi asked to which clans
they belonged. The spokestroll of the group
replied:
"I am a Liar; the lady is a Truthful; the
young gentletroll is a mixed breed."
To which clans did each belong?
2.7. [a] After Heidi (see Exercise 2.5) had
been on the island for a few days, she realized
that each individual's name was woven into his
or her clothing and easy to see.
When Heidi was introduced to Winken,
Blinken, and Finken, she asked her usual
question. The replies she received were:
Winken: I am an Alternator.
Blinken: Winken is a Liar.
Finken: Winken is a Truthful.
If no two of the three belonged to the same
clan, to which did each belong?
2.8. [h] When Heidi (see Exercise 2.5) asked
her usual question to Hocus, Pocus, Crocus,
and Cloy, only Hocus and Pocus replied.
"I am a Truthful; Pocus is a Liar; Crocus
is a Liar; Cloy is a Truthful," was Hocus'
response.
"No, Hocus is a Liar; I am a Truthful;
Crocus is an Alternator; Cloy is a Truthful,"
corrected Pocus.
To which clan did Hocus, Pocus, and
Crocus belong?
2.9. Each year, the leaders of the three clans
(see Exercise 2.5) participate in an athletic
competition to determine which of the three
should govern the island for the new year.
After this year's competition, the following
statements were made by the competitors:
Ange: I came in second; Benge came in
last.
Benge: Ange came in last; Conge won.
Conge: I came in last; Ange won.
What was the order of finish in the
competition?
2.10. [a] The inhabitants of the planet Logos
invented robots that respond to any question,
provided that the proper coin is dropped in
the slot in the robot's chest. Actually, there are
two different model robots. One variety,
Mendibles, respond truthfully when a coin
made of flacus is inserted, but lie if any other
kind of coin is used. The second model,
Lawbakes, respond truthfully to coins made of
flacus or grenjo, but lie whenever a telic coin
is inserted.
On the first visit of Jake Cooke (Captain of
the star cruiser Ganymede) to Logos, he was
confronted by two robots. The Captain found
a supply of coins that were all obviously made
of the same material, although he had no idea
what the material in question was.
Inserting a coin in the first robot. Captain
Cooke asked, "What model robot are you?"
"A Mendible," was the reply.
"And what kind of coin is this?" asked the
Captain, inserting another coin.
"Grenjo," answered the robot.
Turning his attention to the second robot.
Captain Cooke inserted a coin and asked what
model the robot was.
"A Lawbake," was the response he received.
Which model was each of the robots, and
what kind of coin was the Captain using ?

64
CHAPTER 2
MM 2.11. [h] On his way to the capital city
of Logos, Captain Cooke (see Exercise 2.10)
came to a fork in the road. Not knowing which
road to take, he decided to ask a passing robot.
However, the Captain only had one coin, and
he was not sure of what material the coin
was made.
How could he determine with absolute
certainty which road to take?
2.12. The Turner Triplets have an annoying
habit—whenever a question is asked of the
three of them, two tell the truth and the
third lies. When I asked them which of them
was born first, they replied as follows:
Werner: Virna was born first.
Virna: I am not the oldest.
Myrna: Werner is the oldest.
Which of the Turner Triplets was bom first?
2.13. [a] When Mr. Cotter entered the early
morning detention room at Benedict Arnold
High School, he found the school's three
biggest troublemakers—Phil Fagin, Ben
Tolliver, and Spike Sykes—sitting like angels
in their seats, with wide grins on their faces.
On the blackboard was a very uncomplement-
ary picture of Mr. C.
"Okay, who's the artist?" the teacher
bellowed.
"I didn't do it," Fagin answered. "I was
out of the room when it happened. Ben did it."
"I'm innocent," protested Tolliver. "The
picture was already on the board when I
arrived. Fagin lied when he said I did it."
"One of us did it," Sykes admitted. "But
I had nothing to do with it. Phil is innocent,
too."
If each of the three students made two
true statements and lied once, who is the
culprit, and who was present when the picture
was drawn?
2.14. Each year the Chamber of Commerce
in Las Vegas, Nevada sponsors a contest to
see who can build the tallest structure using
only a deck of old playing cards. This year's
contest was very exciting. Four contestants,
Ace Arlington, Jack Johnstone, Emery Queen,
and King Collins had surpassed the 3-foot
high mark. Suddenly, an underground nuclear
test in Los Alamos shook the building and
caused all of the card houses to collapse. It
was decided to award the prize to the person
who was in the lead at the time that the
Shockwaves struck, but there was a lack of
consensus as to who that person was.
r^Slfr,
Ace said: I beat King, but Queen's house
was taller than mine.
Jack said: My building was taller than Ace's,
but Queen won.
Queen said: Jack won. I came in second.
King said: Ace won. I came in second.
Fortunately, they were able to determine the
winner by viewing the videotape replay. It
turned out that each of the four made one
correct and one incorrect statement.

SOLVE IT WITH LOGIC
65
What was the order of finish of the four
contestants ?
M 2.15. d] [a] On Wednesday, May 5, 1976,
the wicked Simon Legrew was murdered at his
home in a Boston suburb. The poHce were
able to place the time of death at between 11:10
and 11:30 pm. They arrested four suspects —
Jeeves, the butler; Fifi, the French maid;
Julia, the cook; and Jessica, Mr. Legrew's
private secretary. Under questioning, the four
suspects made the following statements:
Jeeves: I didn't do it. Jessica did it. Mr.
Legrew was blackmailing Jessica. Fifi and I
were watching television together from 10:10
PM until 12:30 am.
Fifi: Tm innocent. Jeeves and I were
watching television together at the time of the
murder. Jessica was being blackmailed. I saw
Jessica speaking to Mr. Legrew at 9:30 pm on
the night of the murder.
Julia: Tm innocent. Jessica was being
blackmailed. Jeeves murdered Mr. Legrew. I saw
Jessica leave the house at 10:00 pm.
Jessica: I did not kill Mr. Legrew. I was
not being blackmailed. I was in Chicago during
the entire night of the murder. Fifi is the
murderess.
Each of the four suspects made two true
statements and told two lies.
Whodunnit?
2.16. The hop-off at the Toad County Frog
Jump featured four participants—Harold
Hawkins' frog Hopalong, Leopold Lohmann's
leaper Longjump, Gerald Glauberman's entry
Gribbit, and Carlotta Cantrell's bull Croaker.
After the contest, the following statements
were made:
Harold: Croaker won. Hopalong came in
second. Hopalong beat Gribbit by five inches.
Gribbit beat Longjump.
Leopold: Longjump won. Croaker beat
Gribbit. Gribbit did better than Hopalong.
Hopalong beat Croaker.
Gerald: Harold took home the prize money
for first place. Gribbit came in second. Long-
jump came in third. Croaker was a distant
last.
Carlotta: Gribbit won. Croaker came in
second. Hopalong beat Longjump. Longjump
finished last.
If each of the four participants made two
true statements and told two lies, what was
the order of finish in the race?
2.17. [a] Probably the first real evidence of
the existence of the Abominable Snowman is
a photograph taken by the renowned
mountaineer Sir Hilary Edmund and his party.
Before the photo was developed. Sir Hilary
was asked to describe the beast. " It was over
7 feet tall, with long white fur, and 6 toes on
each foot," replied Edmund.
" Him a liar," objected famed Sherpa guide,
Nenzing Torkay, grinning. "The Snowman
has no fur at all, is under 5 feet tall, and
has hooves."
The other three members of the expedition
also disagreed in their descriptions.
Monte Everest© said that the Snowman was
6 feet tall, had long white fur, and hooves.
Matty Home claimed the Snowman was
over 7 feet tall, had brown fur, and had 5
toes on each foot.
And Snowsov "Killer" Manjaro estimated

66
CHAPTER 2
the beast's height at under 5 feet and
attributed brown fur and 6-toed feet to the
creature.
These discrepancies in the descriptions were
probably due to the fact that the beast was
sighted during a severe snowstorm. When the
picture was developed, it proved that each of
the five mountain climbers was correct about
exactly one aspect of the Snowman's
appearance.
What does an Abominable Snowman look
like? (Adapted from [10], Problem 16)
Matching Problems
2.18. [a] Marvin Mandlebaum shares a
medical office with three women—Claudine
Coates, Melissa Mavis, and Rosalinde Rowe.
One of the four doctors is a cardiologist;
another is a gastroenterologist; a third is an
endocrinologist; and the fourth is a hema-
tologist.
According to the rumors that I heard when
I was a patient in the hospital.
Dr. Coates, the cardiologist, and the
gastroenterologist were all sorority sisters in
college.
The cardiologist and the gastroenterologist
were once legally married to each other.
Dr. Mavis and the hematologist are engaged
to be married to each other.
Marvin and his girlfriend play bridge each
week with the gastroenterologist and her
husband.
If exactly three of these statements are true,
what is each person's medical specialty?
2.19. [a] Five playmates, Jason, Danny,
Ethan, Mark, and Mindy, had birthdays
within two weeks of each other, so their
parents decided to give them a joint party. In
order to cut down on presents, it was decided
that each child should bring one gift. The first
to arrive would give his or her present to the
second; the second would give his or her
present to the third; and so on; the fifth giving
his or her present to the first.
After the party, the children were asked
what they had received.
Jason said, "I received the present Ethan
brought."
Ethan said, "I received Mark's present."
Mindy said, "Jason received my present."
Mark said, " I received Ethan's present."
Danny said, " I gave my present to Ethan."
As is to be expected with children, not all
of these statements are accurate. In fact,
exactly three are correct.
From whom did each child receive his or
her present?
M 2.20. [s] [a] Four men were asked about
their yearly incomes. Their names are Earl,
Moe, Luis, and Randy and their professions
are architect, carpenter, plumber, and mason
(not necessarily in that order). Each made two
statements; but the only statements whose
correctness can be depended on are statements
in which the speaker specifically names his
own profession.
Other statements may or may not be true.
Earl: The plumber makes three times as
much as the carpenter. The architect makes
more money than I do.
Moe: The carpenter makes more money
than the plumber. Luis is either the mason or
the architect.
Luis: I make more than the architect. The
carpenter makes less than each of the others.
Randy: The plumber makes twice as much
as the carpenter. I make more than the mason.
Match each person with his profession.
-¥ 2.21. Three women (Abby, Janice, Linda)
and two men (Martin, Roberto) are a singer,
a dancer, a comic, a television writer, and a
theatrical agent, although not necessarily in
that order.
Abby said: I'm not the comic. The writer
and the dancer are happily married. The

SOLVE IT WITH LOGIC
67
singer and the agent are engaged to be married.
Janice said: The singer is my cousin. The
writer and the dancer are siblings. The comic
and the agent share an apartment.
Linda said: I am not the writer. The singer
and the agent hate each other. The dancer
and the comic frequently work together.
Martin said: The singer owes me $10. The
writer and the dancer are not related and have
never met. The comic and the agent are next-
door neighbors in an apartment house.
Roberto said: The singer saved my life once.
The agent lives alone in an old mansion. The
dancer and the comic have never met.
If the only certainty is that every statement
in which an individual alludes to his or her
own profession is true, who is who?
Connectives and Arguments
2.22. [a] Amy, Betty, Carmen, and Dee spent
all their money at the corner store. One girl
spent her dime on candy; another spent her
quarter to buy a ball; a third bought a
coloring book for thirty-five cents; and the
fourth spent forty cents on two comic books.
Each paid for her purchase with the exact
change. Upon leaving the store, the girls made
the following true statements.
Amy said: "If I had a quarter, then so
did Betty. Betty had a nickel if and only if
Carmen did. If I had a dime, then so did Dee."
Betty said: "If I had a quarter, then so did
Amy. If I had a nickel, then so did Dee.''
Carmen said: " If I had a dime, then so did
Amy. If Amy did not have a quarter, then
neither did I.''
Dee said: "If I had a quarter, then so did
Carmen. Betty had a dime. None of us had
any pennies.''
What did each girl buy?
M M 2.23. [h] Prove that if there are more
red cards in the top half of an ordinary deck
of playing cards than there are black cards in
the bottom half, then somewhere in the deck
there are seven consecutive cards of the same
color. ([41], Vol. 26, p. 167)
♦ ♦ 2.24. [D [a] When the first astronaut to
visit the planet Mars returned to Earth, he
was asked to describe the inhabitants of the
" red planet." Still suffering from the effects of
interplanetary travel, he answered in the
following correct, but confusing, manner.
"It is not true that if Martians are green
then they either have three heads or else they
cannot fly, unless it is also true that they are
green if and only if they can fly and that they
do not have three heads."
Assuming that all Martians look alike and
that they have at least one of the three
characteristics referred to:
Do Martians have three heads?
Are they green?
Can they fly?
^ 2.25. [a] Melvin Muddle was having
trouble with his car, so he brought it to his
mechanic. After a careful examination of the
engine, the mechanic said, " It's hard to be
certain, but either it's true that if the spark
plugs and the points are O.K. then you need
a new distributor cap, or else it's true that if
the points arc O.K. but you need new spark

68
CHAPTER 2
plugs then your distributor cap is fine, but
not both."
Assuming that at least one of the three items
mentioned by the mechanic needs replacing,
what should Melvin replace?
2.26. Stanley Plumb, Bing Cherry, and
Walter Mellin decided to form a recording
group—The Three Fruits. Not only do they
all have wonderful singing voices, but each
also plays either the guitar or the banjo.
If Stanley and Walter can both play the
guitar, then so can Bing. If Bing cannot play
the guitar, then Walter can; but if Walter
plays the banjo, then Stanley does not. Either
Stanley or Bing, but not both, can play the
guitar.
Only one of the three can play both the
banjo and the guitar. Which one?
2.27. [a] Nat E. Newcomb owns two suits,
one blue and one brown. Whenever he wears
his blue suit and a blue shirt, he also wears a
blue tie. He always wears either a blue suit or
white socks. He never wears the blue suit
unless he is also wearing either a blue shirt or
white socks. Whenever he wears white socks,
he also wears a blue shirt.
Today, Nat is wearing a gold tie. What else
is he wearing?
2.28. [h] Archie, Brian, and Joaquim own a
Ford, a Chevrolet, and a Chrysler, but not
necessarily in that order. One car is blue, one
is green, and the third is brown.
Archie does not own the Ford, and his car
is not blue.
If the Chevy does not belong to Archie,
then it is green.
If the blue car either is the Ford or belongs
to Brian, then the Chrysler is green.
If the Chevy is either green or brown, then
Brian does not own the Ford.
Identify each person's car in terms of make
and color.
M 2.29. [a] When Luigi opened the box from
his new toy airplane, a piece of paper fell out.
It read as follows:
If this plane is able to fly more than 25
feet high and if this is model #25, then four
penlight batteries are required. But if this is
model # 25, then either the plane is not able
to fly more than 25 feet high or else four
penlight batteries are not required. On the
other hand, if this plane is able to fly more
than 25 feet high, then, if four batteries are
required, this must be model #25.
Assuming that either it was model #25 or
that four penlight batteries were required,
could Luigi's plane fly more than 25 feet high?
M 2.30. [h] [a] Pamela Potter's pease
porridge is putrid provided that Pablo Picasso
painted potted palms. Either Pablo Picasso
painted potted palms, or Peter Piper did not
pick a peck of pickled peppers. There are two
possibilities: Either Peter Piper picked a peck
of pickled peppers or else it is impossible
that both Pablo Picasso did not paint potted
palms and that Pamela Potter's pease porridge
is not putrid.
Is Pamela's porridge putrid?
M 2.31. [h] Unless Manfred Mensa is not
telepathic, it is inconceivable that he is not a
marvelous magician. Either electronic devices
are not used in Manfred's act, or else, if mind
reading is not a form of magic, then Manfred
is not a marvelous magician. If Manfred is not

SOLVE IT WITH LOGIC
69
wealthy, then either he is not a marvelous
magician or else he uses electronic devices in
his act. Either mind reading is not a form of
magic, or else Manfred is not telepathic.
If Manfred is telepathic, is he wealthy?
M M 2.32 [h] [a] If life begins at eighty and
Attila the Hun died at the age of seventy-nine,
then Attila the Hun never lived and he was
not reincarnated as a snake. However, if life
does not begin at eighty and the emotional
development of primates parallels that of
reptiles, then either Attila the Hun never lived
or the female viper is not deadlier than the
male. It is well known that unless the emotional
development of primates does not parallel that
of reptiles, either Attila the Hun never lived or
else he died at seventy-nine; it is also true that
the emotional development of primates
parallels that of reptiles provided that the
female viper is deadlier than the male. On
the other hand, if the female viper is not
deadlier than the male, then it is not possible
that either life begins at eighty or that Attila
the Hun died at seventy-nine.
Attila the Hun lived.
Did he die at the age of seventy-nine?
Does the emotional development of primates
parallel that of reptiles? Was Attila the Hun
reincarnated as a snake?

From Words to Equations:
Algebraic Recreations
&
Interest in recreational mathematics goes back many centuries. For
example, Problem 3.2 below is taken from the Greek Anthology, a
collection of number problems in epigrammatic form, that was
assembled by Metrodorus in about the year 500. According to
Eves [14], there is reason to believe that many of these problems are
even older in origin. At the time that they were originally proposed,
most such problems were solved by trial and error. In fact, there are
probably many people today who would still use the hit and miss
approach. However, these problems are essentially algebra problems,
which can be set up and solved quite simply, as we shall see.
The word algebra is derived from the title of a book on the subject
(as it existed at the time), Hisdb al-jabr w*al-muqdbalah, which was
written by the Persian mathematician Al-Khowarizmi (c. 825). The
term al-jabr apparently meant "restoring," as in restoring the balance
between the two sides of an equation by subtracting from one side
what has been subtracted from another.
The beginnings of the development of symbolic algebraic notation
can probably be traced back to Diophantus of Alexandria (c. 275),
who introduced abbreviations for some of the quantities and
operations that occur most frequently. However, it was not until the sixteenth
and seventeenth centuries that the symbolism of today began to come
into its own.
Algebra, as it applies to recreational problem solving, has two
aspects—the translation of the information of the problem into one or
more equations (or inequalities), and the algebraic solution of these
equations.
70

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
71
The translation begins when one or more variables—x, y, a, r, or
any other letters we deem suitable—are introduced to represent the
unknown quantities of the problem.
After appropriate variables have been chosen, they can be used in
equations to express the given relationships between the unknown
quantities of the problem. These equations may be of any degree and
may contain any number of variables. In most of the recreational
problems considered in this chapter, the equations obtained are
solvable by the techniques of elementary algebra. These techniques,
when applicable, enable us to do more than just find a solution. We
will be able to determine whether or not solutions exist; and, if there
are solutions, we will be able to find all of them. In particular, if there
is a unique solution (that is, only one solution), then the method will
indicate that this is the case.
You are probably familiar with the basic algebraic techniques. If this
is not the case, you might want to look at Appendix A for a review of
some of these techniques before reading on.
The main difficulties, if there are any, in solving a verbal algebraic
problem, usually lie in setting up the equations, that is, in translating
from English into symbols. It is this aspect of algebraic problems that
is briefly considered in this chapter.
The following Sample Problems are used to illustrate our discussion.
Try them first, then read on.
Problem 3.1
The magician said to a volunteer from the audience:
Pick a number, but don't tell me what it is.
Add 15 to it.
Multiply your answer by 3.
Subtract 9.
Divide by 3.
Subtract 8.
Now tell me your answer.
"Thirty-two," replied the volunteer.
At this point, the magician immediately guessed the number that
the volunteer had originally chosen.
What was this number, and how did the magician know so quickly ?
SAMPLE
PROBLEMS

72
CHAPTER 3
Problem 3.2
Demochares has lived one-fourth of his life as a boy, one-fifth as a
youth, one-third as a man, and has spent thirteen years in his dotage.
How old is the gentleman?
Problem 3.3
Bob, Mary, and Sue all have birthdays on the same day. Bob's present
age is two years less than the sum of Sue's and Mary's present ages.
In five years. Bob will be twice as old as Mary will be then. Two
years ago, Mary was one-half as old as Sue was.
How old is each of them now?
Problem 3.4
In a Grand Prix automobile race. Speedy Ryder averaged 110 miles
per hour for the first half of the course and 130 miles per hour for the
second half. Cannonball Carter maintained a constant speed of 120
miles per hour throughout the race.
Who won the race?
Problem 3.5
Alphonse and Gaston, working together, can complete a job in thirty-
six minutes. Working alone, Alphonse would require half an hour
more than Gaston would to complete the same job.
How long would it take Alphonse, working alone, to complete the
job?
INTRODUCING VARIABLES
Before we consider the solutions of these problems, let us examine
the steps followed in translating a verbal problem into one or more
equations. As with any problem presented in words, you must first
read it carefully to make sure you understand exactly what information
is given and what you are required to find.
The next step is to introduce letters, such as x, y, z, to represent
unknown quantities. These letters are called variables. It is frequently
a good idea to choose variables that are reminiscent of the quantities
they represent. For example, we may use n to represent a wumber,
b to represent Bob's age, a to represent a number of apples, and so on.
Many algebra problems involve more than one unknown. You may
introduce a variable for each unknown quantity. Be careful, though.

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
73
not to use the same letter to represent two different quantities, unless
you know that these quantities are equal to each other. Sometimes you
may be able to reduce the number of variables by making use of known
relationships to represent some of the unknowns in terms of others.
For example, if b represents Bob's current age, then Bob's age five
years from now may be represented by 6 H- 5, his age ten years ago
may be represented by b - 10, and his age when he is twice as old as
he is now may be represented by 2b.
three less than twice the number.
the product of the number and two less than the
[a] Each of the following refers to two quantities. Introduce a
variable to represent the first of these quantities, and then represent
the second quantity in terms of this same variable.
(a) A number; the number obtained by adding 15 to the original
number.
(b) A number;
(c) A number;
number.
(d) Demochares' age; one-fifth of his age.
(e) The number of minutes it takes Alphonse to do a job; how
much—what portion—of the job Alphonse can do in one minute.
Each of the following refers to three quantities. Introduce variables
to represent the first two of these quantities, and then represent the
third quantity in terms of the variables used for the first two.
(a) Sue's age; Mary's age; the sum of Sue's and Mary's ages.
(b) Sue's age; Mary's age; the difference between their ages five
years from now. (Sue is older than Mary.)
(c) Sue's age; Mary's age; half of Sue's age two years ago.
(d) The number of minutes it takes Alphonse to do the job; the
number of minutes it takes Gaston to do the job; how much—
what portion—of the job they can do in one minute if they
work together.
PRACTICE
PROBLEMS
3.A
We are now ready to present the solution of Problem 3.1 (page 71).
The trick lies in using algebra to keep track of the volunteer's number
at each step in the procedure.
If you were not able to solve this problem previously, try it
again now.

74
CHAPTER 3
Solution of
Problem 3.1
Since we don't know what number the volunteer chose, we represent
it by some variable, say n. Then we can keep track of the volunteer's
number at each stage by making a chart.
Magician says
Pick a number
Add 15
Multiply by 3
Subtraa 9
Divide by 3
Subtract 8
Volunteer obtains
n
« + 15
3(n+ 15) = 3« + 45
3n + 36
n + 12
« -1- 4
The final answer obtained by the volunteer is w + 4, four more than
the number originally chosen. Therefore, all the magician has to do
is subtract four from the answer announced by the volunteer. That is,
« + 4 = 32
n = 28.
After each of the unknown quantities of a problem has been
represented in terms of variables, the next step is to use the clues of
the problem to symbolically express the relationship(s) between these
quantities by means of equations or inequalities. Note that some of
these relationships may have already been used in representing the
quantities symbolically.
As an illustration consider Problem 3.2 (page 72).
Solution of
Problem 3.2
Try it again now if you haven't already solved it.
The key relationship between the quantities in this problem is:
Years as a boy H- Years as a youth + Years as a man
+ Years in his dotage = Demochares' current age.
We can proceed in two ways:
Let d = Demochares' current
age
b = the number of years he
spent as a boy
Let d = Demochares' age. Then,
from the statement of the
problem, we may represent the
number of years he spent as a

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
75
y = the number of years he
spent as a youth
m = the number of years he
spent as a man.
Then, the given relationships
of the problem yield the
following equations
b = ld
y = \d
m = \d
b -^ y + m + \3 = d.
This is a system of four
equations in four unknowns. We
may solve it by substituting for
6, jv, and m in the fourth
equation to obtain
\d-^\d + \d-^ \3 = d.
Solving this equation,
\3 = d-ld-\d-^,d
Hence, d = 60.
Ud-
boy by \dy the number of years
he spent as a youth by \d, and
the number of years he spent as
a man by \d.
The key relationship of the
problem may be expressed as the
equation
\d + \d + \d+ 13= d.
Solving this equation,
13 = d-\d
Hence, d = 60.
\d-\d = ^,d.
Note that both approaches to the problem eventually come down to
the solution of the same equation: \d + \d -\- \d + \3 = d. Since this
equation has a unique solution, we know that there is a unique solution
to the problem. That is, Demochares was 60, he spent 15 years as a
boy, 12 years as a youth, 20 years as a man, and 13 years in his dotage.
Observe that although the second approach to Problem 3.2 is slightly
shorter, both approaches to the problem are equally correct. The
difference lies in the work saved by representing all of the unknowns
in terms of one variable (using three of the relationships of the problem
in the process). But, in any problem, if you do not see how to cut down
on the number of variables you must introduce, there is still nothing to
worry about. You cannot go wrong by introducing too many variables;
the worst that can happen is that the resulting system of equations will
be more complicated than it might otherwise have been.
Sometimes, it is not completely clear how to express some of the
relationships between the variables in terms of equations. One
technique that is often helpful is to write each sentence in words and.

76
CHAPTER 3
underneath, to translate each part into its symbolic form. For example,
suppose B and M represent Bob's and Mary's current ages respectively;
then the sentence " In five years. Bob will be twice as old as Mary will
be" may be translated as follows:
In five years, Bob will be twice as old as Mary will be
B -f 5 = 2 (M -f 5)
Observe that the words will be, was, are, is, and so on are often
represented by " = ."
If you are unsure whether you need to add, subtract, or perform
some other arithmetical operation, try substituting specific values for
the unknowns to see what effect each operation would have. For
example, in translating "Bob's age is two less than the sum of Sue's
and Mary's ages,"
Bob's age is two less than the sum of Sue's and Mary's ages
B
S + M
on which side of the equation does the "-2" for "two less than"
belong? Try an example: If Sue is five and Mary is three, then Bob
clearly would be six.
6 = (5 + 3) - 2
Thus B= S ^ M -2
(not 6-2 = 5 + 3)
(rather than B - 2 = S -^ M).
PRACTICE
PROBLEMS
3.B
1. [a] Write the equation or inequality obtained by translating each of
the following sentences into symbols:
(a) A number is equal to three less than its square.
(b) The square of a number is twelve more than the number.
(c) One more than a number is less than or equal to one-fourth of
the square of the number.
(d) Felippe can seal more envelopes in an hour than can Sean and
Misha together.
2. Write the equation or inequality obtained by translating each of the
following sentences into symbols:
(a) The difference between the cube of a number and the number is
six less than ten times the number.
(b) Two years ago, Mary was half as old as Sue was.
(c) Six apples and five pears weigh more than seven pears and five
apples.
(d) One half of Pedro's sheep added to one-third of his horses is
equal in number to half of his horses minus half his sheep.

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
77
We are now ready to solve Problem 3.3 (page 72). Try it
again yourself, before reading on.
Let B, M, and 5 represent the present ages of Bob, Mary, and Sue
respectively. We have already expressed the first two clues of the
problem as equations:
B= S ^ M-2
B ^5 = 2(M -f 5)
The third clue gives:
Two years ago, Mary was one-half as old as Sue was
M -2 = i S-2
M -2 = l(S - 2)
We have obtained a system of three equations in three unknowns. This
system can be solved either by the method of substitution or by the
method of elimination. We present both methods of solution below
(also see Appendix A).
Method of substitution
(\) B= S -^ M - 2
(2) B -f 5 = 2(M -f 5)
(3) M -2 = l(S - 2)
Using (1) to substitute for B
in (2), we obtain
(5 + M - 2) -f 5 = 2(M -f 5).
Simplifying,
5-fM-f3 = 2M-flO
(4) 5 = M -f 7.
Using (4) to substitute for S,
(3) becomes
M -2 = K(M -f 7) - 2).
Simplifying
M - 2 = KA^ + 5)
2M - 4 = M -f 5
Al = 9.
Now, using (4),
5 = M-f7 = 9-f7=16
Method of elimination
We rewrite the equation in the
form
(10
(20
(30
B
B
- S
- M =
2M =
S -^2M =
-
5
2
2
Subtract (T) from (20- The
system becomes
(10 ^-
(2'0
(30 -
- 5
5
54
- M
- M
2M
Now add (2'0 to
(10 B-
(2'0
(3'0
From (3
(2'0,
- 5
5
'0,
- M
- M
M
M =
= -2
= 7
= 2
(30:
= -2
= 7
= 9
9. Then from
5 = Af + 7 = 9 + 7=16.
Solution of
Problem 3.3

78
CHAPTER 3
and using (1),
B = 5 + M-2
= 16 + 9 - 2 = 23.
Finally, from (T),
B = S + M-2
= 16 + 9 - 2 = 23.
Thus, Bob's age is 23, Sue's is 16, and Mary's is 9.
Note again that the algebraic method allows us to conclude that the
solution is unique.
Both Problems 3.2 and 3.3 are examples of age problems, one of the
many types of verbal problems. Most types of verbal problems
primarily require the ability to set up and solve equations. Some types,
however, require a bit of outside knowledge such as elementary
formulas from geometry or the laws of motion. For example, uniform
motion problems (often called rate-time-distance problems) require
knowledge of the fact that rate = distance divided by time or
distance = rate multiplied by time.
Problem 3.4 (page 72) is such a problem. If you haven't
already solved it, try it again now.
Solution of
Problem 3.4
Chances are that if you have never solved a problem like this before,
then your first impression was that the race was a tie.
But before you jump to conclusions, consider a slight variation of
the problem. Suppose Cannonball maintained a constant speed of 120
miles per hour throughout the race but that Speedy averaged 60 miles
per hour for the first half of the course and 180 miles per hour for the
second half. Would it still seem that the race was a tie? Suppose
further that the course was 120 miles long. Then Cannonball finished
after one hour; but, at the end of that first hour. Speedy was just
finishing the first half of the course. Clearly the race was not a tie.
But why not? Certainly 120 is the average of 60 and 180. The answer
is that Speedy spent more time at the lower speed than he did at the
higher speed. To finish the first half of the course at 60 miles per hour
took Speedy 1 hour; but to finish the second half at 180 miles per
hour required only \ hour = 20 minutes. Thus Speedy spent more time
at the lower speed than at the higher speed, and so his average speed
was less than 120 miles per hour.
The same type of reasoning applies in general. The amount of time
each driver spent at each speed is important. This becomes clear if we
ask ourselves what it means for one racer to win the race. It means that

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS 79
he took less time to complete the course than the other racer did.
This observation enables us to attack the problem algebraically, in
terms of inequalities.
In the Speedy versus Cannonball race, suppose the total distance for
the course is d miles.
Let fc represent the number of hours it takes Cannonball to
complete the course,
and
let fs represent the number of hours it takes Speedy.
Then, using the formula
or, equivalently
rate • time = distance
distance
time =
rate
we find that
d
To find fs> we consider the two parts of the course during which
Speedy's speed was constant. That is,
h = tx + t2^
where t^ is the time required for Speedy to complete the first half of
the course and ^2 is the time required for him to complete the second
half. Since each half is dl2 miles,
^ dl2 d _ ^/2 _ d
^'~TT0~220 ^'~T30~260
and
^s - ^ + ^2 - 220 "^ 260 ~ 20 \11 "^ 13/ ~ 20 \143/ ~ 715'
The question now is, who completed the course first? Is
fs < tc or is h > tc or does Ts = ^c ?
Since
1 6 6
= <[
120 720 715'
d 6d
<c
120 715
That is, tc < tsi and so Cannonball won the race.

80 CHAPTER 3
There are several important aspects of the solution above. The first
is that we used fc, ?s> ^ and ^2 to represent four different times. It
wasn't necessary to do this: We could have chosen c, 5, b, and e (or
any four other letters) to represent CannonbalFs time, Speedy's total
time, Speedy's beginning time (for the first half of the course) and
Speedy's ending time (for the second half); but we decided to use the
mnemonic t to help indicate that we were dealing with periods of time.
As we could not represent four different times by the exact same
symbol t, we introduced the subscripts c, s, 1, 2 so that we could
differentiate between the different times. The symbol tc is read
"f sub c" or simply "f c"; the symbol t^ is read "f sub 1" or
"f one"; and so on. In general, the decision about whether to use
subscripts or different letters to represent the unknowns of a problem
is up to you; we recommend that subscripts be used when several of
the unknowns in a problem represent the same kinds of quantities
(several time periods, several distances, etc.).
The second important observation in the solution of Problem 3.4
relates to the notion of dimensional consistency. When we
introduced the variable tc to represent Cannonball's time, why did we let
tc represent a number of hours rather than a number of minutes or
seconds? The answer is that since distance was being measured in
miles and rate was being measured in miles per hour, then time should
be measured in hours; that is,
distance miles
time = = —-— = hours.
rate miles/hour
In general, in any equation involving physical measurement, you
must make sure to maintain consistency. If you measure time in hours
in one part of a problem, then you must measure it in hours
throughout. This may require conversion from one unit to another
(for example, writing \ hour rather than 20 minutes, or | mile rather
than 2640 feet). Similarly, as in the solution above, the units used for
one quantity may determine the units to be used for another. For
example, if rate is measured in feet per second, then distance should
be measured in feet and time should be measured in seconds.
Dimensional consistency can sometimes be used to help check the
reasonableness of an equation. Thus, if you write an equation in which
hours are added to miles, then something is wrong. Only quantities
expressed in units of the same type can be added to each other.
Similarly, the units on both sides of an equation must balance. For
example, in the formula rate • time = distance
miles
hour
hours = miles.

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
81
SO that both sides of the equation are expressed in terms of miles.
Thus, if the units on one side of an equation result in an expression
in terms of miles, then the other side of the equation must also be a
number of miles and not, for example, a number of miles per hour.
The solution of Problem 3.4 also leads to a third consideration.
What was the key to being able to attack the problem algebraically?
In this problem, it was the realization that what had to be compared
were the times of the two racers. In general, you must ask yourself
the question, "What kind of quantities must I consider in this problem
in order to get a meaningful equation or inequality?"
For example, in attacking Problem 3.5, there is a tendency to
proceed as follows:
Let A = the amount of time (in minutes) that it would take
Alphon^e, working alone, to complete the job
and
let G = the amount of time it would take Gaston alone.
Then
A + G = 36 minutes.
But is this really meaningful? Is the sum of Alphonse's time and
Gaston's time equal to the time it takes the two of them together?
Certainly not. Together, it should take them less time than it takes
either of them working alone. In this problem, then, it is not
meaningful to add times. But in that case, how can we approach the
problem ? What kinds of quantities should we compare ? How can we
represent the fact that they are working together? How about adding
the amount of work they each do in the same time period ?
Think about this suggestion. Then, if you haven't already
solved Problem 3.5 (page 72), try it again.
Upon rereading the problem, we observe that the information in one
of the clues is stated in terms of minutes and that the information in
the other clue is stated in terms of hours. To maintain dimensional
consistency, we must choose one of these units and stick to it. Which
we choose is a matter of personal preference. The problem can be solved
either way, and each approach has its own advantages:
Let A = the number of minutes
it would take Alphonse, working
alone, to complete the job; and
let G = the number of minutes
it would take Gaston working
alone.
Let a = the number of hours it
would take Alphonse, working
alone, to complete the job; and
let g = the number of hours it
would take Gaston working
alone.
Solution of
Problem 3.5

82
CHAPTER 3
How much of the job could
Alphonse do in one minute?
Since it takes him A minutes
to do the whole job, he does — of
A
the job in one minute.
Similarly, Gaston does — job
G
in one minute.
Together, they do —h — job
A \j
in a minute. (Note, here it is
meaningful to add.)
Therefore, in 36 minutes they
do 36
[^^h]'^'"'-
But we are told that they do
the whole job in 36 minutes.
Thus
or, equivalently,
V 1 1 1
1
How much of the job could
Alphonse do in one hour?
Since it takes him a hours to
do the whole job, he does - of
a
the job in one hour.
Similarly, Gaston does - job
g
in one hour.
Together, they do - + - job in
a g
an hour. (Note, here it is
meaningful to add.)
Observe that 36 minutes = \
hour.
Therefore, in 36 minutes, they
do three-fifths of what they can
do in an hour—that is.
Since we are told that they do
the whole job in 36 minutes.
We are also told that Alphonse, working alone, requires one-half hour
(30 minutes) more than Gaston does. Hence,
(2) /I = G -f 30.
Using (2) to substitute for A in
(1), we get
1 1
G -f 30 "^ G ^
1
36"
Combining over a common
denominator,
G G+ 30
G{G + 30)
G{G -f 30)
2G -f 30
G{G -f 30)
1
36
1
36'
(20 a^g + l
Using (2') to substitute for A in
(!')> we get
sW + i
= 1.
Simplifying,
5 \2^ + 1
2g . 2g
= 1
+
5 \g{2g + 1) g{2g

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
83
Cross-multiplying,
36(2G -f 30) = G{G + 30)
12G -f 1080 = G^ + 30G
G" - 42G - 1080 = 0.
This may be solved by factoring
or by using the quadratic
formula:
(G - 60)(G -f 18) = 0
G = 60, G = - 18
(Or, using the formula,
G =
42 ± V(42)' - 4 • 1 • (-
2
42 ±^1764 +4320
2
42 ± ^/6084
2
42 ±78
2
42 + 78 42 - 78
or
- 1080)
2
60 or
18.)
3 / 4^+1
5 \g{2g + 1)/ ^
3(4^ H- 1) = 5g{2g H- 1)
12^ H- 3 = 10^^ H- 5g
10^ - 7^ - 3 = 0.
This may be solved by factoring
or by using the quadratic
formula:
(^- l)(10^H-3) = 0
^ = 1>^ = -r'o
(Or, using the formula.
g
=
1±JV-
2
4 • 10
• 10
7 ± ^/49 + 120
20
7±^/^i69
20
7 ± 13
20
7+13 7
or -
20
1 or -r\.)
- 13
20
(-3)
The negative answer is extraneous (it satisfies the equations but
does not satisfy the conditions of the problem) since a negative time
does not make sense. It may therefore be discarded.
Thus G = 60, and so
A = G -¥30 = 90
That is, Alphonse, working
alone, could do the job in 90
minutes.
Thus ^ = 1, and so
a=g + \=\\
That is, Alphonse, working
alone, could do the job in an
hour and a half.
Note that both methods lead to the same answer (90 minutes =
\\ hours).

84
CHAPTER 3
Each method has some minor disadvantages. The first results in an
equation with relatively large numbers, so that factoring or taking the
necessary square root is more difficult; the second requires more work
with fractions, but, overall, seems to result in easier computations. But
whichever units you would choose, the mathematical steps in the
solution of the problem are basically the same.
THE CHAPTER IN RETROSPECT
In this chapter, you have gained experience in translating from words
to equations. Even though your experience was won by solving
problems of recreational mathematics, the techniques of this chapter
are useful in many fields. Algebraic problems arise in many settings-
business, economics, social science, and others (see, for example.
Exercise 3.6). Once you recognize that a problem is algebraic in nature,
the basic steps for solving it are the same, regardless of the setting:
1. Represent the unknowns by variables.
2. Set up the equations relating these variables. Be consistent with
the dimensions (in fact, the dimensions can sometimes give a clue as
to how to set up the equation).
3. Solve the equations or show that there is no solution.
In this chapter, we have considered only problems that can be solved
by elementary methods. Sometimes a problem can be set up
algebraically, but the method of solution is not obvious or not known.
Then more mathematics might be needed (some examples of this type
will be considered in Chapter 4). In general, though, the algebraic
approach is very useful.
Exercises
Problems Involving One Variable
3.1. The union election was quite close.
Two-fifths of the votes went to Millie, five-
twelfths of the votes went to Jacob, and the
remaining 33 votes were cast for Harriet.
How many votes were cast in all?
3.2. [s] [a] To celebrate their wedding
anniversary, Derek and Lucy decided to spend the

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
85
night out on the town. After dinner, which
incidentally cost them one-third of the money
they had brought with them, they purchased
two theater tickets at $9.80 each. After the
show, they spent one-fourth of their remaining
bankroll for a taxi ride around the city. They
then went to a night club, where their total
tab came to $23.10. The cab fare home cost
them one-half of their remaining money. After
they had tipped the driver a dollar, they had
$4.10 left.
How much did their celebration cost them ?
3.3. A local charity had set a goal of $70,000
in its fund drive. This morning, when I was
solicited for a donation, I asked how the drive
was doing. I was told that one-third of the
amount that has already been collected is
equal to three-fifths of the amount still needed.
How much money is still needed?
^ 3.4. [g [a] Mildred, Ethel, Samuel, and
Leonard purchased a bag of candy which they
divided as follows:
First, Samuel took one candy plus one-third
of the candy remaining. Next, Mildred took
one candy plus one-third of the candy then
remaining. Then Ethel took one candy plus
one-third of the candy then remaining.
Leonard received the remainder of the candy.
If Mildred and Ethel together received
seven pieces of candy more than Samuel, how
many pieces of candy did Leonard receive ?
3.5. [s][a] Two mendicants decided to spend
their day's proceeds to buy some Muscatelle.
Tom was able to buy seven bottles, while
Don could afford only five. Just as they were
about to begin. Mack showed up and asked
to join them. He offered to pay $8.40. If all
three men shared the wine equally, how much
of the $8.40 should Tom receive?
3.6. Ann Mann has decided to join her
brothers Stan and Dan in a business venture.
They plan to form a trucking company, with
twelve trucks in all. Stan is contributing seven
vans and Dan five. Ann has agreed to pay her
brothers $120,000 in compensation. How
should Stan and Dan divide this money?
Quadratic Equations in One Variable
3.7. [h] Farmer Gray bought a number of
sacks of wheat seed, three times as many sacks
of com seed, and six times as many sacks of
soybean seed (as of wheat). Coincidentally,
the price per sack of each kind of seed was
the number of sacks of that kind of seed that
Farmer Gray bought.
If Farmer Gray spent a total of $184, how
many sacks of seed did he buy ?
3.8. [a] Farmer Gray owns two rectangular
fields of the same area. One is 700 yards longer
than it is wide; the second is 450 yards shorter
than the first, and 400 yards wide.
What are the dimensions of each field ?
3.9. The floor of a square room is to be tiled
according to the pattern in Figure 3.1.
FIGURE 3.1
Both white sections are themselves square
and the larger white square has exactly eight
tiles more on each side than the smaller one.
If 1000 white tiles are needed in all, how
many gray tiles are required?

86
CHAPTER 3
M 3.10. [s] [a] Assuming that there is a
unique number x that may be expressed in the
following form:
1 + VI +^\ + ••• ,
where all roots are positive square roots and
where ... means "and so on, ad infinitum,"
find X.
♦ 3.11. [h] [a] Assuming that there exists a
unique number x that may be expressed in the
following form:
X = 2 -f
15
2+15
2+15
2 + •••
find the value of x.
Systems of Equations
3.12. [a] If a towel and a washcloth together
cost $2.20, and the towel costs $1.20 more
than the washcloth, how much does the
washcloth cost?
3.13. [a] Whenever a cattle drive crosses Solly
Shepherd's land, he charges a toll of 10^ per
riderless animal and 25^ per cowboy and horse
combination. Yesterday, as a drive was
passing, Solly counted a total of 4248 legs
(including riders, horses, and cattle) and 1078
heads. How much money did Solly collect?
3.14. [h] [a] Calvin Carpenter needed screws
of three different sizes. He bought the same
number of screws of each size at $.03, $.04,
and $.05 per screw respectively.
Wendy Woodall spent the same amount of
money as Calvin, but divided her money
equally among screws of the three sizes. In all,
Wendy purchased six more screws than Calvin
did. How much money did they each spend?
3.15. [s] [a] At the weekly meeting of Plumps
Anonymous, only six members—Ada,
Brendan, Corinne, Darryl, Eva, and Floyd—
flunked the weigh-in; they had not lost the
prescribed number of pounds.
Ada and Brendan together tipped the scales
at 647 lb.
Brendan and Corinne together weighed
675 lb.
The sum of Corinne's and Darryl's weights
was 599 lb.
DarryFs and Eva's weights totaled 583 lb.
If Floyd weighed 370 lb and the sum of
the weights of all six was 1927 lb, how much
did each weigh?
3.16. Because each of the people in Exercise
3.15 had not lost enough weight, each had to
pay a fine. Ada and Brendan together paid
$1.80; Brendan and Corinne paid $4.30;
Corinne and Darryl paid $4.00; Donald and
Eva paid $3.80; Eva and Floyd paid $3.10;
Ada and Eva paid $2.80.
How much was each fined?
3.17. [a] Minski, Pinski, Dubinski, and their
wives went shopping. Each of the three couples
spent a total of $51. The sum of the amounts
spent by Minski, Pinski, and the wife of
Dubinski was $43. Minski, Dubinski, and the
wife of Pinski together spent $93. Pinski,
Dubinski, and the wife of Minski spent $81.
How much did each spend?

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
87
3.18. [a] The paper currency of the Kingdom
of Bonoria bears the pictures of the country's
monarchs. The one-bonor note carries the
picture of Queen Griselda the Good. Other
notes not exceeding 100 bonors bear the
pictures of King Randolph the Rotten, Queen
Carrie the Charming, Queen Bonita the
Beautiful, King Gerald the Gross, King Waldo the
Wicked, and King Hilary the Hairy.
Together the notes on which Bonita's and
Carrie's pictures appear are worth 102 bonors.
One Gerald, Waldo, and Randolph together
give 73 bonors. Hilary and Bonita give 22
bonors. Hilary and Carrie give 120 bonors.
Hilary, Gerald, and Randolph give 43 bonors.
And Carrie, Waldo, and Randolph give 168
bonors.
On what size note does Waldo's picture
appear?
^ 3.19. [a] The silver currency of the
Kingdom of Bonoria (see Exercise 3.18) consists of
glomeks, nindars, and morms. Four glomeks
are equal in value to seven nindars; and one
glomek and one nindar together are worth
thirty-three morms.
On my last visit to Bonoria, I entered a
bank, handed the teller some glomeks and
nindars, and asked him to change them into
morms.
"Do you think that I am a magician?" he
replied. (Bonorians are noted for their warped
sense of humor.) " Well, let's see," he
continued. " If you had twice as many glomeks,
I could give you 120 morms; and if you had
twice as many nindars I could give you 114
morms."
How many morms did he give me?
on baubles, then she would have had as many
baubles as she had bangles and beads together;
had she spent all of her reserve money on
bangles, then she would have had twice as
many bangles as she had baubles and beads;
had she spent all of her reserve money on
beads, then she would have had four times as
many beads as baubles and bangles together.
How many of each item did she purchase ?
3.21. [a] The Fastbinder Family Circle
decided to have a catered get-together this year.
The cost was to be divided equally among all
of the individuals attending. If everyone that
was expected had shown up, the cost per
person would have been $8. As it was, eight
people canceled out at the last minute, so each
person's share had to be raised to $9.
How many people attended the get-
together?
3.22. [a] When the wagon train left Fort
McConnell and headed West, each of the
wagons contained the same number of people.
By the time the caravan had crossed the
Mississippi River, eight wagons and four
people had unfortunately been lost, so that
there were now exactly two more people per
wagon than there were when they started.
Crossing the Plains, they lost three more
wagons and eight more people during an
Indian attack. But one baby was born on the
way, so that, when the wagon train arrived in
California, each wagon carried three more
people than it did at first.
How many of the people who left Fort
McConnell with the train survived the trip ?
3.20. A merchant borrowed 120 dinars to go
into business selling baubles, bangles, and
beads. Each item she purchased cost her 1
dinar. She stocked some of each item, but kept
some of the borrowed money in reserve for
emergencies (such as bribing the palace guard,
etc.). Had she spent all of her reserve money
Age Problems
3.23. My age is the difference between twice
my age four years hence and twice my age
four years ago. How old am I ?
3.24. Two years ago, I was three times as old

88
CHAPTER 3
as I was eight years before I was half as old as
I will be six years from now. How old am I ?
3.25, [s] [a] Sue Ling is three times as old as
Chin Lee was when Sue was as old as Chin is
now. When Chin is as old as Sue is now. Sue
will be 56. How old are Sue and Chin now?
^ 3.26. [h] [a] Mutt is half as old as Jeflf will
be when Mutt is twice as old as Jeflf was when
Mutt was half as old as Jeflf is now. In five years,
the sum of Mutt's and JeflPs ages will be 100.
How old are Mutt and Jeflf now?
-¥ 3.27. When Erica was two years old,
Leroy was four times as old as Miriam.
When Miriam was twice as old as Erica, Leroy
was three times as old as Miriam. How old was
Erica when Leroy was twice as old as Miriam ?
-¥ 3.28. [h] \a\ The butcher is twice as old as
he was when the baker was half as old as was the
candlestick maker. The candlestick maker is
twice as old as the butcher was when the
baker was half as old as the butcher is now.
Twelve years ago the baker was half as old as
the butcher will be twelve years from now.
How old is each of them now?
::^V<^^^
-¥ 3.29. My nephew Seymour has four pet
hamsters—Norma, Izzy, Susie, and John—
whose ages he measures in months. Norma is
three times as old as John. In three months
Izzy will be as old as the other three combined.
In five months Izzy will be three times as old
as Susie will be. When Izzy is twice as old as
Norma, Susie will be one and one half times as
old as John.
How old are the hamsters now?
3.30. Lee is two years older than Brenda.
The product of their ages is 575. How old are
they?
Uniform Motion Problems
3.3L [a] When the tortoise raced the hare,
the former maintained a constant pace of one
mile per hour throughout the race, while the
latter, being overconfident, wasted much time
and averaged only a half mile per hour for the
first half of the course.
How fast must the hare run over the second
half of the course in order to win ?
3.32. [s] [a] Two trains are traveling toward
each other on adjacent tracks. The first train is
270 feet long and is moving at 45 miles per
hour. The second train is 192 feet long and
is moving at 60 miles per hour.
How much time elapses from the moment
the trains first meet until they completely pass
each other? (Note that 1 mile per hour =
88/60 feet per second.)
3.33. [h][a] At 8:00 am a train leaves
Topeka for Santa Fe and another train leaves
Santa Fe for Topeka. The trains maintain
constant speeds with no stops. The first train

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
89
requires five hours to complete the trip, and the
second train requires seven hours.
At what time do the trains pass each other ?
^ 3.34. \a\ a stagecoach leaves Deadwood
headed for Tombstone at the same instant
that another coach leaves Tombstone for
Deadwood. They meet along the way at a point
which is 24 miles closer to Deadwood than it
is to Tombstone. Since each driver wants to
return to the town from which he left, the
drivers trade coaches. That is, the driver from
Deadwood takes the coach from Tombstone
on to Deadwood, completing the trip in nine
more hours; and the driver from Tombstone
takes the coach from Deadwood on to
Tombstone, arriving 16 hours after the changeover.
Assuming that each coach travels at a
constant speed throughout the entire trip
(ignore the few minutes required for the
changeover of the drivers), what is the distance
from Deadwood to Tombstone?
3.35. [h] At 8:00 am, a train leaves
Washington, D.C., bound for Miami. At the same
moment, another train leaves Miami headed
for Washington.
Each train maintains a uniform speed
throughout its trip and makes no stops until
it arrives at its destination.
If the trains pass each other at 5:00 pm, and
the train from Miami arrives in Washington
at 11:00 PM, at what time does the train from
Washington arrive in Miami?
3.36. [a] a ferry is crossing the English
Channel from Dover to Calais at a speed of
6 knots (nautical miles per hour). A second
boat is crossing from Calais to Dover at a speed
of 10 knots.
How far apart are the two boats one hour
before they pass each other?
3.37. [h] [a] Two driver less trains race
toward each other on the same track; one is
traveling at a constant speed of 60 miles per
hour and the other at 70 miles per hour.
A fly starts at the front of one train and
flies toward the second train. When he reaches
it, he turns around and flies back to the first
train, whence he turns around, and so on. The
fly maintains a constant speed of 110 miles
per hour, losing no time in changing direction.
If the trains are now 65 miles apart, how
far will the fly fly before he is crushed when
the trains collide?
3.38. [s] [a] Two cyclists are holding a race
on a straight track. They begin at opposite
ends of the track. Each races to his opponent's
end, turns around, races back, turns around,
and so on. The winner is the cyclist who first
overtakes his opponent.
The race begins. The cyclists first pass each
other (in opposite directions) at a point 150 yd
from the northern end of the track. They

90
CHAPTER 3
next pass each other (in opposite directions)
at a point 100 yd from the southern end.
If they maintain constant speeds and lose
no time in turning
(a) How long is the track?
(b) Who will win the race?
(c) At what point will the winner overtake
the loser?
M ^ 3.39. [h] [a] One of the ski lifts at Snow-
flake Mountain climbs the mountain in a line
that parallels one of the ski runs. Both the run
and the lift are two miles long, and a chair
passes the starting point of the lift every 10
seconds.
A skier starts her run just as a chair arrives
at the top of the lift and another chair starts
back down. She arrives at the bottom just as a
chair is starting up the mountain and another
chair is completing its descent. Counting these
two chairs and the two at the start of her run,
she sees 97 chairs on their way up the
mountain and 61 chairs on their way down.
If the chair that started down the slope at the
same time that the skier did is still on its way
down when she reaches the bottom of the
slope, what is her average speed?
3.40. Jeremiah Jordan wanted to test the
power of his new outboard motor. He took his
boat out on the Mississippi River and noted
that, with the throttle wide open, he could go
5 miles upstream in 9 minutes and he could
go 5 miles downstream in 7 minutes.
How many minutes would it take him to go
5 miles if there were no current ?
3.41. [h] [a] As Rosalind Rowe rowed
upstream on the Raritan River, she noticed a
piece of driftwood floating by. She continued
rowing upstream for half an hour, when she
suddenly decided that her husband would find
the driftwood quite decorative. Rosalind
quickly turned around, headed downstream
after the ornament, and caught up to it 2 miles
downstream of the spot at which she originally
saw it.
Assuming that Rosalind maintained a
constant rate of rowing throughout, how fast was
the current in the river?
M 3.42. [s] [a] Just as Liz and Irv stepped
off the express train, they could hear the local
pulling in to the upper platform of the station.
They quickly ran up the escalator, Irv taking
three steps for each two that Liz took.
Unfortunately, the doors of the local closed
just as Irv reached the top step of the escalator.
Since they had to wait for another local to
arrive, Liz and Irv had time to reflect about
the escalator. Liz remarked that it had taken
her 24 steps to reach the top; Irv noted that
it had taken him 30. Assuming that Liz and Irv
each climbed the escalator at a constant rate,
how many steps would be visible if the
escalator stopped running?
M 3.43. [h] [a] The absentminded professor
walked up the escalator in the department
store at a constant rate of one step per second.
Just as he reached the top, he realized that he
had left his briefcase on the counter where he
had just purchased a pocket calculator.
Fortunately, no one else was on the up escalator,
so the professor decided to run down. He
managed three steps per second on his descent.
After he retrieved his briefcase, the professor

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
91
decided to repeat the procedure. He noted that
he took 18 steps on the way up and 90 steps on
the way down.
PIAMTS
How many steps of the escalator are visible
at one time?
M M 3.44. [s] [a] At 4:15 PM, things looked
bleak for supercilious Pilious Flog and his
three traveling companions. According to the
terms of their bet, they must all arrive at
Winchester Cathedral by 6:00 pm; but they
were still 19.3 kilometers from their
destination and could only sustain a jog of 4
kilometers per hour.
Just then. Flog flagged down a passing
motorcycle, which had room enough for one
passenger. Unfortunately, none of the four
men could drive, but Pilious was able to enlist
the aid of the motorcyclist. Three of the men
would continue jogging while the motorcyclist
would drive one of the men part of the way to
the cathedral. That man would continue on
foot, and the driver would return for another
of the voyagers, to whom he would then give
a ride until they caught up with the first
passenger, and so on.
By the time these arrangements were
completed, it was 4:20 pm.
Assuming that the motorcycle travels at the
rate of 56 kilometers per hour and that the
pedestrians continue to jog at 4 kilometers per
hour, did Pilious and his friends have enough
time to win their bet? Explain.
M 3.45. [h] [a] Harriet Harrier was in a
hurry. She had to reach 34th St. and 6th Ave.
as soon as possible. When she entered the
Continental Avenue station both an EE Local
and an F express train were waiting. This left
Harriet several options:
1. She could take the F all the way to
34th St.; but the F train spends two minutes
in the Queens Plaza station waiting for a
local to arrive.
2. She could take the EE all the way; but
then she would still have one station to go at
the time that the F train would be arriving
at 34th St.
3. She could take the EE to Queens Plaza
and change for the F. Actually the F would
reach Queens Plaza by the time the EE had
only gone 10 stations; but, owing to the F's
stopover at Queens Plaza, the EE would
arrive just in time for Harriet to change for
the F.
4. She could take the F to Queens Plaza
and arrive just in time to catch an earlier
EE. This last route is the fastest, saving
one full minute over the time required to
travel the entire route by F train.
Assuming that the trains in question
maintain uniform speeds (in terms of stations per
minute) except for the two minute period when
the F train is stationary, how many stations
are there between Continental Avenue and
34th St? And how long will Harriet's trip take
if she travels by the fastest route?
M 3.46. \a\ Comm. Dale E. Muter travels to
and from work by train. His train arrives at
his hometown station at 6 pm each evening,
and his wife always arrives promptly at this
time to pick him up. One day, Comm. Muter
left work early and arrived at the station at
5 PM. Not wishing to disturb his wife, he
started to walk home along the route that she

92
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always drove. When he was one-quarter of the
way home, he met his wife. They proceeded
home at their usual driving speed and arrived
home 12 minutes earlier than usual.
If the Muters live 12 miles from the station,
how fast was Comm. Muter walking?
Work Problems
3.47. The representative from Leonardo and
Winslow Paint While You Wait Co. came to
give me an estimate on the cost of painting
my house.
"If we work with our regular crew," she
said, "we can have the job done in eight
hours. But if we hire three extra people, it will
be done in six hours."
Assuming that all the painters work at the
same rate, how long would it take one of them
working alone to paint the house?
3.48. Arnold, Hank, and Wilhelm are the
three mailmen in the town of Littlewood. On
an average day, if Arnold is not working. Hank
and Wilhelm, working together, can sort the
mail in 36 minutes; if Hank is not working,
Arnold and Wilhelm can sort the mail in an
hour; if Wilhelm is not working, Arnold and
Hank can sort the mail in half an hour.
(a) [a] How long will it take all three of
the men, working together, to sort the mail
on an average day?
working alone, to sort the mail on an average
day?
3.49. [a] Bertram Bovine has two drinking
troughs for his cows. Both troughs have
vertical sides a foot deep, but one is rectangular
in shape and the other is circular.
Half of Bertram's herd of cows can drain
the rectangular trough in 24 minutes, and the
other half of the herd can drain the circular
trough in 28 minutes.
If both troughs are filled and half the herd
drinks from each, in how many minutes will
the water in the circular trough be twice as
deep as the water in the rectangular trough?
3.50. [a] If, on the average, a chicken and a
half can lay an egg and a half in a day and a
half, how many average eggs are laid by seven
average chickens in seven average days?
(b) If Arnold is on vacation and Wilhelm
calls in sick, how long will it take Hank,
-¥ 3.51. [a] Peter Piper and his wife Pepper
have a vegetable garden and a fruit orchard.
Working together they can collect the
harvest from the garden in 3 hours, whereas
Pepper, working alone, requires 12 hours.
Furthermore, together they can harvest the
orchard in 2 hours, whereas Peter, alone, takes
10 hours.
It would seem that to harvest both the
garden and the orchard, they should first spend 3
hours in the garden and then 2 hours in the
orchard—a total of 5 hours in all. However,
Peter is much more skillful at picking vegeta-

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
93
bles and Pepper is better at picking fruit, so
that they can save time by having Peter work
in the garden and Pepper work in the orchard
until one of them finishes. That person would
then help the other.
How much time can they save in this
manner?
M -¥ 3.52. [h] Gilbert Greenfield owns a
pasture on which the grass has been cut to a
uniform height of two inches. The grass in the
pasture grows uniformly at a constant rate.
In addition to owning the pasture, Mr.
Greenfield also owns a cow, a horse, and a
sheep.
The grass existing and growing in the
pasture is sufficient to feed all three animals
grazing together for 20 days. It would feed the
cow and the horse alone for 25 days; the cow
and the sheep alone for 33| days; and the horse
and sheep alone for 50 days.
How long would it sustain
(a) [a] The cow alone ?
(b) The horse alone?
(c) [a] The sheep alone ?
Miscellaneous Problems
3.53. {a\
At a college class reunion from dear old N.Y.U.
I met fifteen former classmates—counting men
and women too.
More than half of them were doctors, and the
rest all practiced law.
Of the former, more were females; of that,
Tm pretty sure.
More abundant still were women who received
a law degree.
And these statements are still truthful, even if
you include me.
If my friend, a noted lawyer, left his wife and
children home.
Can you draw any conclusion about me from
this short poem?
3.54. [h] Leroy Jefferson went shopping. He
bought some shirts and some pairs of slacks,
paying a whole number of dollars for each.
Actually, the price per shirt, in dollars, was
equal to the number of shirts that Leroy
bought; and the price per pair of slacks was
equal to the number of pairs of slacks he
purchased. If Leroy spent $39 more on slacks than
he did on shirts, and if three times the number
of shirts that Leroy bought is equal to seven
more than the number of pairs of slacks he
purchased, how many items of each type did
Leroy buy?
3.55. [s] [a] Chefjeffhad just finished
pouring one pint of vinegar into one large cruet and
one pint of oil into another when he was called

94
CHAPTER 3
to the telephone. During his absence, his
child, Julia, decided to take over. She removed
one level tablespoon of oil from the cruet of
oil and carefully mixed it into the cruet of
vinegar. Then, before any separation could
take place, she took one tablespoon from the
mixture and mixed it into the cruet of oil.
When Jeff returned, did he find more oil in
the vinegar or more vinegar in the oil ? (A pint
is equivalent to 32 tablespoons.)
3.56. Malcolm Milquetoast owns a dairy
farm. The milk that he gets from his cows
contains a cream content of 4 percent. In
order to produce skimmed milk, he must
reduce the amount of cream to 1 percent.
(a) \a\ How much cream must be removed
from 100 gallons of natural milk to produce
skimmed milk?
(b) How many gallons of natural milk must
farmer Milquetoast collect in order to produce
100 gallons of skimmed milk?
3.57. A grocer has 3 lb of almonds which
she purchased for $1.05 per pound and 5 lb
of cashew nuts for which she paid $6.25. If
she can purchase peanuts at $.80 per pound,
how many pounds of peanuts should she buy
in order to be able to produce a mixture which
she can sell for $1.30 per pound to give her
a total profit of $7.00?
3.58. [a] The Blubber boys were slightly
overweight. The four together tipped the
scales at 1160 lb. Whelan, the oldest and the
heaviest of the lot, weighed 20 lb more than
Wally, who in turn weighed 20 lb more than
Bubba, who in turn weighed 20 lb more than
Bobby, the baby of the family.
Three of the boys were married. Beulah
weighs as much as her husband. Barbra's
husband weighs twice as much as Barbra
does; Belinda's husband weighs one and one
half times as much as Belinda.
Altogether, the seven Blubbers weigh 1780
lb. Who is married to whom ? And how much
does each Blubber weigh?
3.59. [a] Using an honest balance scale, you
find that a tweezer and a bar of soap weigh
as much as three combs.
A bar of soap weighs the same as three
toothbrushes and a comb.
Three tweezers weigh the same as a comb
and a toothbrush.
How many toothbrushes are needed to
balance a bar of soap?
3.60. [h] Jeff, Cookie, Chuck, Pearl, and
Bruce went shopping together. Pearl spent the
most money, and Bruce spent the least, Jeff
spent more than Chuck, who in turn spent
more than Cookie. The total amounts spent
by each pair of individuals were $7, $10, $11,
$14, $15, $16, $17, $18, $20, and $24 (for
example, Jeff and Pearl together spent $24).
How much did each person spend?
-¥ 3.61. [h] [a] A dishonest grocer owns a
balance scale for which one arm is slightly
longer than the other. Whenever she purchases

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
95
produce from the local farmers, she places the
produce in the pan on the shorter arm of the
scale, so that the produce seems lighter than
it actually is. When the grocer sells produce,
she naturally places it in the pan which is
further from the fulcrum (the point of
balance.)
Figure 3.2 shows how the grocer weighs six
tomatoes and a sack of green beans when she
buys them, and then how she weighs the
tomatoes and green beans separately when she
sells them.
FIGURE 3.2
The principle is that the force exerted by an
object on a balance scale is the product of the
weight of that object and the length of the arm
on which the pan is located. Thus, for
example, if an object balances with a 1 lb
weight on an honest scale, then that same
object will balance with a 2 lb weight if the
pan on which the object is located is twice as
far from the fulcrum as the other pan which
contains the known weight.
(a) What is the ratio of the arm lengths of
the scale?
(b) How much do the six tomatoes weigh?
(c) How much does the bag of beans weigh?
Ancient (and Not Quite as Ancient)
Problems
The problems below can be found in the
following sources, all of which predate the
eighteenth century.
1. The Rhind Papyrus (c. 1550 bce).
2. Greek Anthology (c. 500) assembled by
Metrodorus.
3. Ganita-Sdra Sangraha (c. 850) by
Mahavlra.
4. Triparty en la science des nombres (1484)
by Chuquet.
5. Algebra Christophori Clavii Bamber-
gensis (1608) by Clavius.
3.62. [a] (Rhind Papyrus) Mass, its 2/3, its
1/2, its 1/7, its whole, it makes 33.
3.63. (Greek Anthology) I am a brazen lion,
a fountain; my spouts are my two eyes, my
mouth, and the flat of my right foot. My right
eye fills a jar in two days [1 day = 12 hours],
my left eye in three, and my foot in four;
my mouth is capable of filling it in six hours.
Tell me how long all four together will take to
fill it.
3.64. (Greek Anthology) This tomb holds
Diophantus. Ah, how great a marvel! The
tomb tells scientifically the measure of his life.
God granted him to be a boy for the sixth part
of his life, and adding a twelfth part to this.
He clothed his cheeks with down; He lit him
the light of wedlock after a seventh part, and
five years after his marriage He granted him a
son. Alas! late-born wretched child; after

96
CHAPTER 3
attaining the measure of half his father's [full]
life, chill Fate took him. After consoling his
grief by this science of number for four years,
Diophantus ended his life.
How old was Diophantus when he died?
3.65. (MahavTra) The mixed price of 9
citrons and 7 fragrant wood-apples is 107;
again, the mixed price of 7 citrons and 9
fragrant wood-apples is 101. O you
arithmetician, tell me quickly the price of a citron and
of a wood-apple here, having distinctly
separated those prices well.
3.66. [a] (Mahavlra) One-fourth of a herd of
camels was seen in the forest; twice the square
root of that herd had gone to the mountain
slopes; three times five camels remained on the
river bank. What is the numerical measure of
that herd of camels ?
3.67. (Chuquet) A merchant visited three
fairs. At the first he doubled his money and
spent $30, at the second he tripled his money
and spent $54, at the third he quadrupled his
money and spent $72, and then had $48 left.
How much money had he at the start?
3.68. [a] (Clavius) If I were to give 7 cents to
each of the beggars at my door I would have
24 cents left. I lack 32 cents of being able to
give them 9 cents apiece. How many beggars
are there, and how much money have I ?
Tricks
Each of the following exercises describes a card
trick or a number guessing trick that is based
on the use of algebra. Your task is to describe
what the magician must do to find the chosen
card or number, and to explain how and why
the trick works,
3.69. [a] Magician: "Think of a number.
Double it. Add 8. Multiply by 5. Subtract 3.
Double the result. Subtract 31. Multiply by 5.
Tell me your result."
Subject: "915."
Magician: "Then the number you started
with is ."
How can the magician quickly find the
number that the subject originally selected?
Explain the trick.
3.70. Magician: "Select a card from this
ordinary deck. Counting the Ace as one. Jack
as 11, Queen as 12 and King as 13, double
the value of the card. Add 3. Multiply by 5.
If the selected card is a club, add 1; if it is a
diamond, add 2; if a heart, add 3; and if it is
a spade, add 4. Now tell me your result."
Subject: "79."
Magician: "The card you selected is
the "
What is the selected card ? And how does the
trick work?
3.71. [h] [a] Magician: "Write an odd
number greater than one and less than ten. Now
write an even number between one and ten.
Subtract the smaller of the two chosen
numbers from the larger to give a third
number. Add the two chosen numbers to give
a fourth number. Multiply the third and fourth
numbers. Tell me your result."
Subject: "39."
Magician: "The numbers you chose
were and ."
Find the selected numbers and explain how
the trick works.
¥ 3.72. [h] [a] Magician: "Here is an
ordinary deck of playing cards. Place it on the
table, face down. Select a number from 10
to 26. Remove the selected number of cards

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
97
from the deck, turn them face up, and replace
them on top of the deck. Shuffle the deck to
distribute the face up cards throughout the
deck. Again remove the selected number of
cards from the top of the deck and place them
in my hands, behind my back. Although I
have no way of knowing how many face up
cards there now are in my hand and how
many remain in the rest of the deck, I shall try
to turn some of the cards in my hands over, so
that the number of face up cards in my hands
is exactly equal to the number of face up cards
in the rest of the deck."
After manipulating the cards behind his
back for a few moments, the magician
produced them and, sure enough, the number
of face up cards in his packet was equal to the
number of face up cards in the rest of the deck.
How was the magician able to produce this
effect?
'^^ 3.73. [s] [a] Magician:" Here is an
ordinary 52 card deck of playing cards. You may
shuffle them if you like. I will turn my back.
While my back is turned, you will make a
number of piles as follows: Turn over the top
card of the deck. If it is a picture card, replace
it in the middle of the deck and turn over the
new top card. Continue doing this until you
obtain a card numbered between 1 (Ace) and
10. When you have such a card, place it face
up on the table. Beginning with the number of
the card on the table, start counting to
yourself until you reach 12. With each count, take
one card from the deck and place it face up on
the top of the pile that you are creating.
When you reach 12, turn the pile over (so that
the cards are face down with the card you
originally selected on top), and begin a new
pile. Thus, for example, if you turn up an 8,
place a card on top of the 8 and silently count
* 9'; then place another card on top of the pile
and count '10'; then place another card and
count Ml'; then place a final card and count
'12.' Then turn the pile over and begin a new
pile. Continue doing this until you do not
have enough cards remaining to complete a
new pile. Give these remaining cards to me."
Subject: "Here they are. Now what?"
Magician: "I will now turn around. I see
that you have six piles and five cards left over.
Remove the top card from each of your piles
and add the values of these cards together.
You will obtain a total of
How was the magician able to determine the
total, and how does the trick work in general ?
M 3.74. [h] [a] Magician: "I have here an
ordinary 52 card deck of playing cards. You
may shufflle them if you like. Now divide the
deck into two equal halves and give me one
half. I'm fanning the cards in my half of the
deck. Select one; look at it; and now put it
back here." (The magician has cut his pile
into two parts; the spectator places his card on
top of the indicated part and then the magician
places the remaining part on top of the
selected card.)
"Now, turn up the top three cards of your
half on the deck. On top of each of these cards,
place as many cards as are necessary to
complete the count to 10. Thus, if you turn up
a 7, you need three more cards to complete the
count—8, 9, and 10. If you turn up a 10 or a
picture card, no cards are needed to complete
the count. (If you do not have enough cards to
complete the count, you may use the top few
cards from my half of the deck.) After you have
completed the count with all three cards, place
the remaining cards from your half of the deck
on top of my half, which I have placed on
the table.
"Next, add up the three cards you have
turned up from your deck. (Count picture
cards as 10.) Remove from the top of the stack
of cards on the table as many cards as the
sum you have obtained. The next card will be
the card you originally selected."
How did the magician know?
♦ 3.75. {a\ Magician: "This is another
variation of the previous trick. Again I have

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an ordinary 52 card deck, which I am shuffling.
I now take the top eleven cards and place
them face down on the table. Choose four of
these cards while I write a prediction on this
piece of paper. Turn the four chosen cards
face up, and I will return the other seven
cards to the bottom of the deck. With each of
the four face up cards, complete the count
to 10 (see Exercise 3.74) by taking cards from
the top of the deck. Now add up the four
numbers (counting picture cards as 10) and
count that many cards from the top of the
deck. Turn up the next card. You will note
it is exactly the card I predicted."
How did the magician know which card to
predict?
-¥ 3.76. [s] The magician handed an ordinary
52 card pack of cards to his first subject and
asked her to remove a packet containing
between one and five cards. He then asked the
subject to pass the remaining cards to a second
subject, who was instructed to remove a packet
of between 10 and 20 cards. The remaining
cards were then returned to the magician.
Magician: "I will now form a pile face up
on the table by removing one card at a time
from my packet. With each card I place on the
table I will count. When I call out the number
equal to the number of cards you originally
selected, say nothing but note the card that I
place on the table at that time. That will be
your card for this trick. To make sure that you
both have adequate opportunities to observe
your cards, I will continue the count until I
reach 26. I will then pick up the pile of cards
from the table and place it, face down, on the
bottom of my packet."
After the magician had completed this
procedure, he divided his packet in half
(leaving the extra card, if there was one, with
the bottom half). The bottom card of the
bottom half was then discarded, and the two
halves were placed side by side on the table—
the top half face down and the bottom half
face up.
Magician: "I will now simultaneously
remove the top card from each pile, and will
continue doing so until you (the second
subject) tell me that the next face up card is
your card. At that point the top face down
card will be yours (the first subject)."
As usual, the magician was correct. How did
he know this would happen?
M 3.77. [h] [a] Magician: "I have here an
ordinary 52 card deck of playing cards which
you may shuffle if you like. Now remove
from the top of the deck a packet containing
between one and 20 cards. Count the number
of cards in the packet, put the packet in your
pocket, and return the remainder of the deck
to me.
"Now I will slowly deal a pile of 20 cards,
one at a time, face up on the table. At the
moment that my pile contains the same
number of cards as your packet, say nothing,
but note the card that is momentarily showing
on the top of my pile.
"Now that I have completed my pile, pick
a number between one and 10."
Subject: "Nine."
Magician: "I shall deal a second pile
containing nine cards, face down on the table. I
now place my original pile face down upon the
second pile and then I place the remaining
cards in my hand face down upon this
combined pile. This places your selected card
somewhere in the middle of the deck, so that
I can have no idea where it is. I will attempt
to find your selected card by handling one
card at a time, without seeing the face, until

FROM WORDS TO EQUATIONS: ALGEBRAIC RECREATIONS
99
the tingle in my fingers tells me that I have
located the correct one.
"Aha! This is your card."
The magician was correct, as usual. How did
he know which card to choose?
3.78. [h] The magician handed an ordinary
52 card deck of playing cards to Jack, the
volunteer from the audience, and asked him to
shuffle the deck, to remove a packet of no more
than 10 cards from the top of the deck, to count
the number of cards in the packet and, using
that number, to locate and memorize the card
that is in that position in the remaining part
of the deck. (If, for example. Jack's packet
contains six cards, then he should note the
sixth card in the remaining portion of the
deck.) The entire deck—with the exception of
Jack's packet—was then returned to the
magician.
Magician: "Now pick a number between
10 and 20."
Jack: "14."
Magician: "I want you to deal a pile of 14
cards, one card at a time, face down on the
table, as follows."
The magician thereupon placed the top card
of his packet face down on the table, counting
" 1." He placed the second card face down on
top of the first, counting "2." And he
continued in this manner until he had a pile of
14 cards. He then picked up this pile, placed
it back on top of the deck and handed the
deck to Jack.
Magician: "Oh, yes, before proceeding,
please return your packet to the top of the
deck."
Jack did as he was told, and, after he had
completed a pile of 14 cards, returned the
remainder of the deck to the magician. The
magician tapped the deck three times, placed
the deck on the table, and asked Jack to name
his card. The subject said, "Jack of hearts."
Magician: " Would you please turn over the
top card of the deck and hold it up to the
audience."
Naturally, the card was the Jack of hearts.
How did the magician know?
3.79. [h] [a] Magician: "I will let you be the
magician for this last trick. Here is an ordinary
52 card deck of playing cards. Hand it to me
and tell me to shuffle it."
Alice: "Here, please shuffle these cards."
Magician: "Now take the cards back, fan
them and tell me to pick a card, look at it,
and replace it in the deck."
Alice: "Pick a card, look at it, and replace
it in the deck."
Magician: "Now ask someone in the
audience for a number between 1 and 10."
Voice from the audience: "6."
Magician: " Count 6 cards off the top of the
deck, placing each one face down on the table
on top of the previous one. Then ask me my
card."
Alice: "What was your card?"
Magician: "The four of diamonds. Now
turn over the next card and it should be the
four of diamonds."
Alice: "Nope. It's the six of hearts."
Magician: "I know what's wrong. You
forgot to blow on the cards. Pick up that packet
of six cards from the table and put it back on
top of the deck. Then ask the audience for a
number between 10 and 20."
Alice (to audience): " Would someone please
pick a number between 10 and 20.
Audience: " 14."
Magician: "Count off 14 cards, face down,
one at a time, as before. Then blow on the next
card, look at it. It should be the diamond four."
Alice: "Nope, wrong again."
Magician: "Can't you do anything right?
Put the packet of 14 cards back on top of the
deck, and I'll show you how it's done. Let me
pick 8. I'll now count off 8 cards, blow on the
deck, and, lo and behold, the next card is the
four of diamonds."
The magician was correct. How was the
trick done?

Solve It With Integers: Some
Topics From Number Theory
The theory of numbers is a field of study that has excited the interest
of mathematicians for centuries. It is primarily concerned with
properties of the positive integers (1, 2, 3, ... )•
In early western and oriental cultures, whole numbers were originally
of interest for ritual and computational purposes. The study of their
properties, however, seems to have begun with the Pythagoreans, a
school of scholars led by Pythagoras (c. 572 bce), of Pythagorean
Theorem fame. They felt that certain whole numbers had mystical
significance.
Over the following two centuries, mathematicians of the time
considered various types of whole numbers: for example, primes (which
we will discuss below), perfect numbers (numbers such as 6 that are
equal to the sum of their proper divisors: 6 = 1+2-1-3), and deficient
numbers (numbers such as 8 that exceed the sum of their proper
divisors: 8 > 1 -f 2 -f 4). Euclid (300 bce) was able to characterize all
even perfect numbers, but to this day it is not known whether or not
any odd perfect numbers exist.
It was not until the 1600s that Fermat, a lawyer and mathematician,
gave importance to the theory of numbers as a field of mathematics in
its own right. His insight gave us many results, such as "Every odd
prime can be expressed as a difference of two squares in exactly one
way." He also left us the famous conjecture that for w, a positive
integer greater than two, x" H- j^" = 2:" has no positive integer solutions
for X, jv, and z.
Many of the problems in the field of number theory are simple to
state but difficult to solve. Some, such as the question of odd perfect
100

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
101
numbers and Fermat's famous conjecture, have in fact withstood all
attempts at solution to this time. However, many of these attempts
have resulted in significant contributions to our mathematical
knowledge.
In this chapter, we are primarily interested in some elementary
aspects of number theory that will be useful in solving recreational
problems. And, of course, we are interested in the types of problems to
which this theory can be applied. Some of these are illustrated in the
following sample problems. (Since these problems relate to
mathematical techniques with which you may as yet be unfamiliar, you may
find some of them quite difficult. You should try them, anyway.
Even if you fail to solve any of them, you will become aware of
what difficulties exist and will better appreciate the techniques when
they are presented.)
Problem 4.1
Heloise Homemaker prepared 84 canapes for a dinner party she was
making. Although she never ate at her own parties, she always
prepared precisely enough hors d'oeuvres for each of her guests to
consume the exact same quantity.
What possibilities are there for the number of guests she has invited?
Problem 4.2
Henry, Eli, and Cornelius and their wives Gertrude, Katherine, and
Anna (not necessarily in that order) each purchased at least one animal
at a farm auction. Coincidentally, each of the six spent an exact number
of dollars for each animal equal to the number of animals he or she
purchased. Henry purchased 23 animals more than Katherine did, and
Eli spent $11 per animal more than Gertrude. Also, each husband
spent a total of $63 more than his wife.
Who was married to whom?
Problem 4.3
(a) Given a 3 gallon jug and a 5 gallon jug (without any markings),
is it possible to get exactly 1 gallon of water from a well? If so, how?
If not, why not?
(b) What if we have only a 4 gallon and a 6 gallon jug? Is it
possible to get 1 gallon of water? If so, how? If not, why not?
Problem 4.4
It was a strange lapse on the part of the bank teller. Evidently, he
misread the check, for he handed out the amount of dollars in cents and
SAMPLE
PROBLEMS

102
CHAPTER 4
the amount of cents in dollars. When the error was pointed out to him,
he became flustered, made an absurd mathematical mistake, and handed
out a dollar, a dime, and a cent more. But the depositor declared that he
was still short of his due. The teller pulled himself together, doubled
the amount he had already given to the depositor (that is, he handed the
depositor an additional amount equal to the total amount he had given
him previously), and so settled the transaction to everyone's satisfaction.
What was the amount called for by the check? ([45], Problem 134)
How are these problems different from the ones we considered in
Chapter 3?
Let us examine Problems 4.1 and 4.2.
Setting Up
Problem 4.1
On the surface, this looks like an algebra problem, and, algebraically,
it may be set up as follows:
Let X = the number of people Heloise could have invited,
and
let y = the number of canapes each person would eat.
Then
xy = 84.
(4.1)
This equation has many solutions; in fact, infinitely many. For
each nonzero value we choose for jc, we can just let y — 84/x and
get a solution. For example, jc = 5, jy = ^^^ is a solution to equation
4.1; so is jc = -2, jv = -42. But neither of these gives a solution to
Problem 4.1. Each guest cannot eat a fractional number i^^') of canapes;
and there cannot be a negative number of guests. That is, in solving
Problem 4.1, we are interested only in solutions of the equation for
which X andjv are positive whole numbers. An obvious solution \%x = 1,
y = 84. Another is jc = 2, jy ^ 42. Are there more solutions and, if so,
how can we be sure of finding all of them?
Before we try to answer this question, let us observe that Problem 4.2
(page 101) leads to a similar situation.
Setting Up
Problem 4.2
This problem is a slightly modified version of one which, according
to Dudeney in his Amusements in Mathematics ([13], pp. 26-27), first
appeared in the 1700s in Ladies Diary.
Again, we may begin to attack the problem algebraically. Let the
number of animals each person bought be represented by the first letter

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
103
of his or her name. By the conditions of the problem, this initial will
also represent the number of dollars he or she spent per animal. Hence,
the total amount spent by each person is represented by the square of
that person's initial. For example, Henry spent }P dollars, (// animals
at H dollars each gives H^ dollars).
Since each man spent $63 more than his wife, we have, for each
couple, the following equation:
63,
(4.2)
where X represents the number of animals the husband purchased andjy
represents the number of animals purchased by his wife. (Note that
equation 4.2 applies to all three couples. As we do not as yet know
who is married to whom, we use x and y rather than specific initials.)
Since each person purchased a whole number of animals, x and y
must be positive integers (whole numbers), and so we are again
interested only in integer solutions of this equation.
This, then, is the basic difference between the problems of Chapter 3
and many of the problems of this chapter. Algebraically, the setup
is similar; but here we are interested only in whole number solutions,
and we need other techniques to find them.
DIOPHANTINE EQUATIONS
We will return shortly to the solutions of Problems 4.1 and 4.2. In
the meantime, let us consider the general type of equation for which
we want integer solutions. Such an equation is called a Diophantine
equation, named after the Greek mathematician, Diophantus of
Alexandria (c. 275).
Definition 4.1 A Diophantine equation is an equation in which
all coefficients are integers and in which we are interested only in
integer solutions.
Any equation with integer coefficients may or may not be considered
as a Diophantine equation, depending on our point of view.
For example, the (linear) equation 3x -\- 5y = 1 has an infinite
number of solutions in general (x = \, y = 0; or x = I, y = -i; and
so on). It also has integer solutions (such as jc = 2, jy = — 1; or x = — 3,
y = 2); therefore, as a Diophantine equation, it has solutions. On the
other hand, the (linear) equation 4x -\- 6y = 1 also has an infinite
number of solutions in general (for example, x = \, 3^ = 0; x = 1,
y = -|). However, we will soon prove that it has no integer solutions;
so, as a Diophantine equation, it has no solutions.

104
CHAPTER 4
As another example, consider the equation x^ -^ y^ = z^* This also
has many solutions in general, such sls x = y = \, z = ^^2; however,
if we consider this as a Diophantine equation—that is, if we are only
interested in integral solutions—then it is known that there are no
solutions other than x = 0, y = z, or y = 0, x = z, or z = 0, y = - x.
Similarly, equations 4.1 and 4.2 have infinitely many solutions in
general; however, to solve the problems from which these equations
arise, we must consider them as Diophantine equations.
Diophantine equations frequently arise in problems such as 4.1 and
4.2 in which noninteger values for the unknowns would be meaningless.
(For example, x might represent a number of animals or a number of
people.)
There are no general techniques that may be used to solve all
Diophantine equations; however, there are methods that may be
applied to certain types of Diophantine equations. Later in the chapter,
we will investigate a method for attacking linear Diophantine equations.
But, for now, we are interested in the quadratic equations 4.1 and 4.2.
In order to consider the solutions of these equations, we need some
information about whole numbers and their properties.
DIVISIBILITY
Let N refer to the set of positive integers; that is, N =
{1,2,3,4,...}.
Definition 4.2 If m and n represent integers, w t^ 0, we say that
m divides n (written symbolically as m\n) if there exists an integer q
such that n = mq. (If m does not divide w, we write m \ w.)t
For example, 2| 10 because 10 = 2 • 5; but 3 | 10 because there is
no integer q such that 3 ■ q = 10.
The following statements are equivalent:
w is a divisor of n
* The claim that the Diophantine equation x" + y" = z" has no positive integral solutions
for any integer n > 2 has become known as Fermat's Last Theorem. Although it is known
to be true for many values of n, including all values of n between 3 and 125,000, in general
it is still an open question as to whether or not it is true for all n > 2. (While reading a
book on Diophantus' work, Fermat noted in the margin that he had a proof of this
result, but that the proof was too lengthy to write in the margin. Unfortunately, no trace
of his proof has been found.) For an interesting discussion of the history of this problem,
see "Fermat's Last Theorem" by H. Edwards, Scientific American, October 1978, pp.
108-122.
t Caution: m|n is not the same as m/n. The latter is sometimes used to denote -.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
105
w is a factor of n
w is a multiple of m
and are sometimes used instead of "m divides w."
Note that "m divides w" refers to the operation of division but is
defined in terms of multiplication.
When we write n = mq, we are factoring n into two factors, each
one being a divisor of n.
Definition 4.2 applies to negative as well as positive integers.
However, for the remainder of this chapter when we speak of the
divisors of a number w, we will mean the positive divisors of n.
The following theorem, which follows directly from the definition of
divisibility, expresses some important properties of divisibility:
Theorem 4.1 Let a, 6, c, and d be integers.
(i) If a|6 and 6|c, then a\c (transitivity of divisibility).
(ii) If a 16, then a\bc.
(iii) If a -\- b = c and if d divides two of a, b and c, then d also divides
the third.
We prove (i) below and leave (ii) and (iii) as an exercise (Practice
Problems 4.A). The remainder of this chapter presents a number of
theorems, usually without proof. Proofs and other related results may
be found in [30] or [47].
Proof of (i):
Since a|6, there is an integer d such that b = ad.
Since b \ c, there is an integer e such that c = be.
Therefore, c = be = (ad)e = a(de).
Since de is an integer, a \ c.
As examples of Theorem 4.1:
3112, and 12184, therefore, by (i), 3|84.
316, therefore, by (ii), 316x for any integer x.
316x and 3115jv by (ii), therefore if ^ + 6x = 15y, then 31 /j by (iii).
1. Suppose k = 6x ■¥ \2y for some integers x and y.
(a) [a] Explain why we can conclude that 21 ^, 31 ^, and 61 k^ even
though we do not know the specific values of x and y.
(b) Show by a counterexample that 12 need not divide k.
2. Prove part (ii) of Theorem 4.1.
3. [a] Prove part (iii) of Theorem 4.1.
PRACTICE
PROBLEMS
4A

106
CHAPTER 4
PRIME NUMBERS
Consider the set N of positive integers. Every integer n in N, other
than 1, has at least two divisors in N—namely, n itself and 1. (Note
that w = wl.)
Definition 4.3 A positive integer/) that has exactly two divisors in N
is called a prime number. (The two divisors are 1 and p.)
Definition 4.4 A positive integer other than 1 that is not a prime
number is called composite.
For example, 2 is a prime; it has two divisors, 1 and 2. Similarly,
3, 5, 7, 11, 13, and 17 are primes.
On the other hand, 4, 6, 8, and 9 are composite.
4 has three divisors: 1, 2, 4
6 has four divisors: 1, 2, 3, 6
8 has four divisors: 1, 2, 4, 8
9 has three divisors: 1, 3, 9
Note that 1 is neither prime nor composite.
Note also that, since a composite number n is not prime, it has at
least one divisor m other than 1 and itself; that is, n = mq where
neither m nor ^ is 1. If w is not prime, then it is composite and
hence has a divisor r other than 1 and itself. By Theorem 4.1 (i),
r is also a divisor of m. If r is not prime, it has a divisor other
than 1 and itself. Etc. Each new divisor is smaller than the previous
one, so the process cannot continue indefinitely. Eventually we must
reach a divisor of n that is a prime number. We thus have the
following theorem.
Theorem 4.2 Every positive integer n other than 1 is divisible by
a prime.
<^ THE INFINITUDE OF PRIMES
How can we find all primes? As a matter of fact, we can't, because there
are infinitely many primes. This fact was already known to Euclid, the
famed geometer (c. 300 bce). His proof is interesting to us because
it uses an idea that we discussed in Chapter 1—if an assumption leads
to a contradiction, then the fact assumed must be false. The proof
also makes use of Theorems 4.1 and 4.2, and proceeds as follows:
There are two possibilities—either there are a finite number of
primes or there are an infinite number. We assume that there are a
finite number, k^ of them and see what happens. Call the k primes
Pn />2 J " ', pk' Let
n = p, ' p2
/>/.+ !•

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
107
By Theorem 4.2, n must be divisible by one of the primes />,. By
the definition of divisibility, the product p^ • p2 • •.. • /)/t is also divisible
by pi. That is,
/>/.!(Pi •/>2 •.../>/.+ 1) andp,|(p, p2 •...•/>/fe),
so that, by Theorem 4.1 (iii),
/>«|1.
This is impossible. Hence, our assumption leads to a contradiction
and the only alternative remaining is that there are infinitely many
primes.
^ THE SIEVE OF ERATOSTHENES
Although it is not possible to find all primes, it is possible to find
all primes less than any given number. One way of doing this is to
use the sieve method, so called because the composite numbers drop
out, leaving only primes in the ** sieve." This method, which was
devised by Eratosthenes of Alexandria (c. 200 bce), proceeds as follows
(we illustrate using n = 50):
Since 1 is neither prime nor composite, we list all integers from
2 to w:
2 3 4 5 6 7
9 10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The first number on the list is 2, a prime. We circle it. The other
multiples of 2 are composite (since they are divisible by 1, 2, and
themselves), so we cross them out.
The first uncircled number remaining is 3. Since it is not a
multiple of 2, it must be prime. We therefore circle 3 and cross out its
multiples, which must be composite. (Note that every third integer
after 3 is a multiple of 3; note also that some multiples of 3, such as 6
and 12, that are also multiples of 2 have already been crossed out.)
This leaves:
@@ A' 5 ^ 1 ^ ^^6 11 J^ 13>4 >5 J^ 17 J^
19 ^ >r ^ 23 X 25 X X ^ 29 :3€r 31 >2' 3^ X 35
^ yi yi y^ ^ M ^ 43 K ^ ^ 47 ^ 49 ^
None of the uncircled numbers remaining can be multiples of 2 or 3
(or they would have been crossed out), and so, by part (ii) of Theorem
4.1, they cannot be multiples of any numbers that are multiples of
2 or 3. Therefore, the first uncircled number remaining, 5, cannot be

108
CHAPTER 4
a multiple of any smaller number (except for 1) and, hence, must be
prime. We circle it and cross out all of its multiples (every fifth number
after 5). (Again note that many of these have already been crossed out.
In fact, as 2 • 5 and 4 • 5 have been crossed out as multiples of 2,
and 3 • 5 has been eliminated as a multiple of 3, 5 • 5 is the first multiple
of 5 that has not been crossed out previously. In general, when using
the sieve method, the first multiple of a new prime p that will not have
been crossed out previously is p ■ p = p^.)
The first uncircled number now remaining, 7, is a prime. We circle
it and cross out its multiples.
We continue in this manner until all numbers on the list have either
been circled or crossed out. (Actually, in our example, once we finish
crossing out the multiples of 7, no further numbers need be crossed out.
Each time we circle a new prime /> > 7, its smallest multiple that has
not already been crossed out, p^, is larger than 50 and hence not on the
list. Thus, all numbers remaining at this point are prime. In general,
once we have eliminated all multiples of primes less than or equal to
^ w, all numbers remaining in the sieve are prime. We will return to
this point shortly.)
In our example, we are left with
® ® ^ ® X ® J^ ^Mf@ X^ @ l^ 0\^@\^
1^^XX(23)X^>2^X^(29)^ (3l)>2'>3' X X
X (37) X >^ 0 (4Vl ^(43) ^ 4^ ^ (47) ^ ^^
So the primes less than 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, and 47.
PRACTICE 1. Use the sieve method to find all primes up to 200.
PROBLEMS
4.B
MORE ABOUT PRIMES
Why are we interested in primes? The main reason is that, in a
multiplicative sense, primes are the building blocks of the integers.
This may be seen from the following theorems:
Theorem 4,3 If a prime divides the product of two positive integers,
then it divides at least one of them. That is, if/> is a prime and p\mn^
then either p\m ot p\n or both.
For example, if/> is a prime and plbO, then—since 60 = 6 • 10—
either /) 16 (p = 2 or 3) or p| 10 (p = 2 or 5). Thus /> = 2, 3, or 5.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
109
Theorem 4.4 (Unique Factorization Theorem) Every positive
integer other than 1 can be factored into prime factors in exactly one
way (except possibly for the order of the factors).
For example,
6 = 2 • 3 (= 3 • 2)
3 = 3
24 = 2-2'2-3(=2
2-3-2 = 2-3-2-2 = 3-2-2-2).
The uniqueness of the theorem may be expressed more precisely
if we use exponents and write the factorization in the form
n = pl^p^^
Pl\
where/>! <p2 < • • • < Py^are distinct primes. In other words, if we write
the primes in increasing order, using exponents to collect powers of the
same prime, then the factorization is unique. For example, instead of
writing 2 • 2 • 2 • 3 • 3, we write 2' • V. Thus,
24 = 2' • V or simply 2' • 3,
300 = 2 • 2 • 3 • 5 • 5 = 2^ • 3 • 5^
and so on.
How do we find the prime factorization of a number? We answer
this question on the left below, while, on the right, we apply the
technique to finding the prime factorization of 44296:
To find the prime factorization
of a number w, we could first try
to divide by 2. If 2|w, then
w = 2 • w, and we then try to
divide m by 2. If 21 w, then
m = 2 ' k, and n = 2^ - k, etc. We
continue until we obtain
w = 2'^ • r, where r is the largest
power of 2 that divides n; that
is, 2 )( t. (Note, if 2 I w, then
r = 0 and t = n.)
We next find the largest
power of 3 that divides t. We
obtain r = 3^ • w, and
yj = 2'' • 3^ • M, where 2 \ u and
3|m.
In this manner, we continue
to factor out the primes 5, 7, 11,
etc. (Note: It is not necessary to
try 4, 6, or any other composite
n = 44296
44296 = 2•22148
= 2 -2 • 11074
= 2 • 2 • 2 • 5537
= 2^•5537
2 )( 5537
3 I 5537
44296 = 2^ • 3« • 5537 = 2^ • 5537
(Note 3^ = 1, and hence is
omitted.)
5 \ 5537
5537 = 7-791
= 7 • 7 • 113
= 72-113

110
CHAPTER 4
number because, if for example,
4|m, then 2 would certainly
divide it but does not.)
Thus, 442% = 2' 'V • 113.
In general, we would now try the next prime, 11, to see if it divides
113, etc. In this example, it is not necessary to do so. This follows
from the general rule that, to determine whether a number is a prime,
it is necessary to test only those primes up to and including the square
root of the number as possible factors of the number. Thus, suppose that
113 could be factored into r • s. Then either r or 5 would have to
be less than or equal to ^113. (If both were greater than ^/ll3,
then rs would have to be greater than w 113 • . 113 = 113.) But we have
already considered all possible prime divisors less than >^/l 13 and have
seen that none of these divides 113; hence, 113 is a prime.
Thus, 44296 = 2^ - V • 113 is the prime factorization of 44296.
Since, by Theorem 4.4, the factorization of 44296 is unique, we
would obtain the same result if we first tried to divide by 7:
44296 = 7 • 6328
and then proceeded to factor 6328; or, if we somehow knew that 14
divides 44296, we could have written
44296 = 14 • 3164
and then have factored 14 and 3164 to obtain
44296 = 2 • 7 • 2^ • 7 • 113
= 2' -V ■ 113.
Although we were able to find the prime factorization of 44296 in a
few steps, the fact is that the process of finding the prime factorization
of a number is sometimes long and tedious, particularly if the number
has a large prime factor.
PRACTICE
PROBLEMS
i4.C
J";:
1. [a] Find the prime factorization of each of the following:
(a) 15120 (b)2183 (c) 409 (d) 72814
2. Find the prime factorization of each of the following:
(a) 3816 (b) 353 (c) 64680 (d) 6006 (e) 44304
(f) your social security number.
Theorem 4.4 is so important in our number system that it is often
called the Fundamental Theorem of Arithmetic. Why? How does
the prime factored form of a number help us? One way is that it gives

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
111
US a method of telling quickly whether a number is a perfect square,
perfect cube, etc.
For example, is 42250000 a perfect square? Note that 42250000 =
24 . 56 . J32^ which is seen by the laws of exponents to be (2^ • 5^ • 13)^.
In general, a number is a perfect square if and only if each exponent
is even when the number is expressed in prime factored form. Similarly,
__-nber is a perfect cube if and only if each exponent in the prime
factored form of the number is divisible by 3.
1. [a] Indicate whether each of the following numbers is a perfect PRACTICE
square and/or a perfect cube. DD^RI PK/IQ
(a) 15120 (b) 46656 (c) 1728. n<yDLIIIVIO
4.D
2. Indicate whether each of the following numbers is a perfect square
and/or a perfect cube.
(a) 784 (b) 353 (c) 8000.
3. [a] (a) For each of the perfect squares in Problem 1, find the square
root.
(b) For each of the perfect cubes in Problem 1, find the cube root.
4. (a) For each of the perfect squares in Problem 2, find the square root.
(b) For each of the perfect cubes in Problem 2, find the cube root.
5. [a] (a) If 1 Ix is a square, what can be said about x?
(b) If 9y is a square, what can be said about y?
6. Find a number that is a perfect fifth power.
7. [a] Give a rule that indicates when a number is a perfect wth power.
Another important consequence of Theorem 4.4 is the following
theorem, which provides us with a method for finding all divisors of a
given number.
Theorem 4.5 If n = p['p2 • • • />!* is the prime factorization of a
positive integer n {p^ < pz < ' <pk are prime numbers), then
(i) every divisor w of w is of the form
where 0 < s^ < rj, 0 < 52 < r^, ..., 0 < 5/^ < r/^.
(Note the possibility that some of the 5,'s may be 0; that is, some
of the /),'s may not appear in the factorization of m.)

112 CHAPTER 4
(ii) The number of divisors of n is (r^ + l)(r2 + \) •' • {rk+ 1).
For example, does 84 divide 392? Writing the prime factorization
of these numbers, we find that
84 = 2^ • 3 • 7 and 392 = 2^ • 1\
From Theorem 4.5, it is now clear that 84 does not divide 392, because
3 appears in the prime factorization of 84 but does not appear in the
392 2^ • V 2-7
prime factorization of 392. (Note: = = , but there
84 2^ • 3 • 7 3
is no 3 in the numerator to cancel the 3 in the denominator.)
Similarly, 112 = 2^ • 7 does not divide 392 = 2^ • V. The exponent
of 2 in the factorization of 112 is too big—it is bigger than 3, the
392 2^ ' V 7
exponent of 2 in 392 ( = = -, but there are not enough 2's
in the numerator to cancel all the 2's in the denominator). On the
other hand, 28 = 2^ • 7 does divide 392 = 2^ • 7^ Each exponent in the
factorization of 28 is less than the corresponding exponent in the
factorization of 392.
Theorem 4.5 even enables us to find the quotient 392 -^ 28:
392 2^ • 7^ 2-7
392 ^ 28 = = = = 14.
28 2^-7 1
Theorem 4.5 can also be used to find all divisors of a number. For
example, to find the divisors of 24, we could proceed by trial and error
(test each positive integer less than or equal to 24 to see if it divides 24),
or we could use Theorem 4.5. Since 24 = 2^ • 3, every divisor must be
expressible in the form 2''3*, where 0 < a < 3, and 0 < 6 < 1.
The divisors may be listed systematically by use of a tree diagram
(see Chapter 1 for a discussion of tree diagrams). This is illustrated in
Figure 4.1.
exponent exponent
divisor
2°3°= 1
203. = 3
2'3« = 2
2'3' = 6
2^ 3° = 4
2^3' = 12
2'3°=8
2'3' = 24
FIGURE 4.1

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
113
Note that part (ii) of Theorem 4.5 tells us that there should be
(3 + 1)(1 + 1) = 8 divisors in all. This may also be seen from the tree
diagram and the use of the Multiplication Principle (see page 19).
1. [a] How many divisors does 6480 have?
2. How many divisors does 44296 have?
3. [a] Find the divisors of each of the following:
(a) 392 (b) 353 (c) 112
4. Find the divisors of each of the following:
(a) 96 (b) 315 (c) 1032 (d) 199
5. [a] (a) Given the square number 2^ • 13"^- 17^, how many divisors
does it have?
(b) What can be said about the parity (oddness or evenness) of the
number of divisors of a perfect square ?
(c) What can be said about the parity of the number of divisors of
a number that is not a perfect square?
PRACTICE
PROBLEMS
4.E
We are now ready to solve the Diophantine equations 4.1 and
4.2 (pages 102 and 103). Try them again if you have not
already solved them.
Since xy = 84 and x and y are positive integers, x and y must be
divisors of 84.
The prime factorization of 84 is
84 = 2^ • 3 • 7.
Hence, by Theorem 4.5, the divisors of 84 are
20 . 30 . 70 ^ l^ 2^ • 3« • 7« = 2, 2^ • 3« • 7« = 4,
20 . 31 . 70 ^ 3^ 2^ • 3^ • 7« = 6, 2^ • 3^ • 7« = 12,
20 . 30 . 71 ^ 7^ 2^ • 3« • 7^ = 14, 2^ • 3« • 7^ = 28,
20 . 31 . 71 ^ 21, 2^ • 3^ • 7^ = 42, 2^ • 3^ • 7^ = 84.
Thus, we can have
X = l,y = 84 x = S4,y=l
X = 2,y = 42 X = 42, y = 2
X = 3^y = 28 X = 28, 3; = 3
X = 4,3; = 21 X = 2\,y = 4
X = 6, y = \4 X = 14, jv = 6
X = 7,y = 12 x=l2,y = 7.
Solution of
Problem 4.1
(Continued)
^-—-o«L v\ V V ''■■/
(SOLUTION] MTikl V -X^
v'/s. v^"^^ /^^TT\7
xi^A—TvU^^^^^'**"

114
CHAPTER 4
Therefore, Heloise could have invited 1, 2, 3, 4, 6, 7, 12, 14, 21, 28,
42, or 84 guests.
Solution of
Problem 4.2
(Continued)
How is the discussion in this chapter relevant to this problem?
Observe that x^ - y^ may be factored into x^ — y^ = {x - y){x + y).
Therefore, equation 4.2 becomes
(X - y){x -^y) = 63.
Since x and y are to be integers, x - y and x ^- y must be a pair of
divisors of 63 whose product is 63. But 63 = 3^ • 7, so, by Theorem
4.5, the only divisors of 63 are
30 . yo _ J 31 . yo _ 3 32 . 70 — 9
30 . 71 _ 7 3^ • 7^ = 21 32 -7' = 63.
Arranging in pairs those numbers whose product is 63, we get
1 and 63 3 and 21 7 and 9.
Since x and y are positive, x — y < x ^ y. Therefore, we have three
possible cases:
(a) X - 3; = 1 (b) X - 3; - 3 (c) X - 3; = 7
x-f3;=63 jc+3; = 21 x^y = ^
In each case, we have two equations in two unknowns, which we
can solve simultaneously to obtain:
(a) x = 32,3; = 31; (b)x= 12,3^ = 9; (c)x = 8,3;=l.
Thus, equation 4.2 has three sets of solutions. The number of
purchases made by each couple in Problem 4.2 must satisfy equation
4.2, where x represents the number of animals purchased by the
husband and y represents the number of animals purchased by his
wife. We can therefore say that the three men (in some order)
purchased 32, 12, and 8 animals and that their wives (in the same order)
purchased 31, 9, and 1 animals.
Since we are told that Henry purchased 23 animals more than did
Katherine and that Eli spent $11 per animal more than Gertrude,
£= G + 11.
The only possibility that is consistent v/ith what we have discovered is
//= 32, K = ^, E= \2, G= \,

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
115
leaving
C =
A = 31.
Thus, Henry (32) is married to Anna (31); Eli (12) is married to
Katherine (9); and Cornelius (8) is married to Gertrude (1). This
completes the solution of Problem 4.2.
LINEAR DIOPHANTINE EQUATIONS
In the solution above, we were fortunate that x^ — y factored easily
into (x - y)(x + jy), so that we were able to use Theorem 4.5. Higher
order Diophantine equations do not behave this nicely in general, and
may not prove to be as easy to solve. However, there is one kind of
Diophantine equation that can always be handled—a linear
Diophantine equation.
A linear Diophantine equation in two unknowns is an equation of
the form
ax + by = c,
(4.3)
where a, 6, and c are known integers, and where we are interested only
in integer solutions—integers x and y which satisfy the equation.
Similarly, a linear Diophantine equation in three unknowns is of the
form
ax + by + cz = d,
etc.
In this chapter we restrict our attention to linear Diophantine
equations in two unknowns.
It follows from Theorem 4.1 that if x and y are integers and if
d is any integer that divides both a and /?, then d\ax, d\by, and so
d\(ax + by). Thus, if equation 4.3 has a solution, then any divisor of a
and b must also divide c. For example, the equation 8x -f 123^= 17
has no solution since the left side is divisible by 4 for any values of
X and y, whereas the right is not divisible by 4.
1. [a] Are there integers x andy such that 3x H- 6y = 5? Explain.
2. Are there integers x and y such that 4x + 6y = 7? Explain.
3. [a] If there are integers x and y such that 3x -\- 6y = k, what can
be said about k?
PRACTICE
PROBLEMS
4.F C^pr)

116
CHAPTER 4
Continuing our discussion, we make the following definitions:
Definition 4.5 A common divisor of a and 6 is a positive integer
that divides both a and b.
For example, 1, 2, 3, 4, 6, and 12 are common divisors of 24 and 36.
Definition 4.6 The greatest common divisor of a and b, denoted
here by gcd(a, b), is the greatest of all the common divisors of a and b.
For example, gcd(24, 36) = 12,
gcd(9, 17) = 1,
gcd(4, 16) = 4.
Definition 4.7 If gcd(a, ^) = 1, then a and b are said to be relatively
prime.
PRACTICE
PROBLEMS
4.G
1. [a] (a) Find the greatest common divisors of each of the following
pairs of numbers:
(i) 16 and 28 (ii) 18 and 26 (iii) 37 and 54 (iv) 15 and 39
(b) Which of the pairs of numbers above are relatively prime?
2. (a) Find the greatest common divisors of each of the following
pairs of numbers:
(i) 34 and 51 (ii) 12 and 35 (iii) 8 and 74 (iv) 113 and 197
(b) Which of these pairs of numbers are relatively prime?
3. [a] (a) If a = 3^ and b = 3\ find gcd(a, b).
(b) If a = p' and b = p', find gcd(a, b).
(c) If a = 3^ • 5^ and 6 = 3^ • 5\ find gcd(a, b).
(d) If the prime factorizations of a and b respectively are
a = p['pl^ ...pl' and b = plpl^ ... pi'
and if, for each i, m, = min{ri, 5,}—that is, the smaller of r, and
5,—find an expression for gcd(a, 6) as a product of powers of
primes.
(e) Use the result in (d) to find gcd( 12600, 15120).
4. (a) If a = 3^ and b = 5\ find gcd(a, b).
(b) If a = 3^ • 5^ and 6 = 2^- 5^ find gcd(a, b).
(c) Find gcd(3816, 15120).
As we observed previously, a necessary condition for equation 4.3
(page 115) to be solvable is that c is divisible by every common divisor
of a and b. In particular, gcd(a, b) must divide c.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
117
Thus, for example, the equation 4x -\- 6y = 1 has no integral
solutions, since 2 = gcd(4, 6) does not divide 1.
It turns out that gcd(a, 6)|c is also a sufficient condition for
equation 4.3 to be solvable. That is, we have the following theorem:
Theorem 4.6 Given integers a, 6, and c, the Diophantine equation
ax -\- by = c is solvable if and only if gcd(a, b) \ c.
We will shortly return to the question of how to find the solutions of
equation 4.3 (when they exist). First let us consider Problem 4.3
(page 101).
If you haven't solved it already, try it again now.
Problem 4.3 is one of a famous category of "decanting" problems. It
is easily found, by trial and error, that one method of obtaining 1 gallon
in part (a) of the problem is as follows (see Figure 4.2).
Fill the 3 gallon jug with water and empty it into the 5 gallon jug.
Refill the 3 gallon jug from the well and use part of its contents to
fill up the 5 gallon jug (which already has 3 gallons in it). This leaves
1 gallon in the 3 gallon jug (as well as 5 gallons in the 5 gallon jug,
which we may spill out). We will see later that this is not the only
solution of the problem.
Part (b) cannot be solved so easily. After many trials, you begin
to feel that 1 gallon of water cannot be obtained using only a 4 gallon
jug and a 6 gallon jug. In every case, there seem to be an even number
of gallons left in each jug. Is this because 4 and 6 are both even
numbers ?
To answer this question, let us reconsider Problem 4.3 in a different
light.
Since the jugs have no markings, if we ever partially fill a jug from
the well, we will have no idea of exactly how much water it contains.
Similarly, if we pour from one jug to the other, leaving them both
partially filled, we will not know exactly how much water is in either.
Finally, as we can never have two partially filled jugs whose exact
contents are known, pouring water from a partially filled jug back into
the well is senseless, because then either both jugs are empty (our
starting position) or one is empty and one is full (a situation that
can be reached directly in one step).
Therefore, the only reasonable moves we could make at any step are:
1. Completely fill an empty jug from the well. (Filling a partially filled
jug is the same as first emptying a partially filled jug and then
filling it completely—a move that we have seen makes no sense.)
Solution of
Problem 4.3
3 gal
u
5 gal
■ U
uy
■ y
urn
uL
FIGUR
E 4.2
(0,0)
(3,0)
(0,3)
(3,3)
(1,5)
(1,0)

118
CHAPTER 4
3 gal
u
u
■
u
y
u
■
u
■
u
5 gal
U
■
u
u
u
■
w
M
u
u
FIGURE 4.3
(0,0)
(0, 5)
(3,2)
(0,2)
(2,0)
(2,5)
(3,4)
(0,4)
(3,1)
(0, 1)
2. Completely empty a full jug into the well.
3. Pour from one jug into the other, completely filling or completely
emptying one jug in the process and leaving the other jug partially
filled.
Let us return to part (a) and set up the problem algebraically as
follows: Count -f 1 every time the 3 gallon jug is filled from the well
and — 1 every time it is emptied into the well. We thus obtain an
integer x which represents that net number of fillings and emptyings
of the 3 gallon jug from the well. (If the 3 gallon jug is filled more
often than it is emptied, then x will be positive; if it is emptied more
often than it is filled, then x will be negative.) In the same manner,
we obtain a number y for the 5 gallon jug. Pouring from one jug to
another, as in a type 3 move above, does not change x and y.
If we obtain 1 gallon by filling (or emptying) the 3 gallon jug x
times and the 5 gallon jug y times, then we must have the equation
3x-^5y= 1,
where 3x H- 5y is the net amount of water that has been removed
from the well. (Note that, if one jug contains 1 gallon, then the other
must be either empty or full. In the latter case, it may be emptied
into the well, leaving only 1 gallon missing from the well.) This is a
Diophantine equation, since x and y must be integers.
Because the coefficients of x and y are small, it is easy to find (by
trial and error) solutions to this equation. One such solution is jc = 2,
y = - \. Another solution is jc = - 3, jy = 2.
Consider the first solution, x = 2, y = - 1. What does this mean?
According to our interpretation of x and y, it means we must fill
the 3 gallon jug from the well twice and empty the 5 gallon jug into
the well once. But how do we fill something twice without emptying it?
And how do we empty something that has nothing in it to begin with?
The answer is simple. We first fill the 3 gallon jug from the well and
then empty it into the 5 gallon jug. (Since we are not emptying it into
the well, this does not affect x.) We then can fill the 3 gallon jug a
second time. We now want to empty the 5 gallon jug into the well. But
it is not full; so we first fill it from the 3 gallon jug (leaving the
desired 1 gallon in the 3 gallon jug), and then we empty the 5 gallon
jug into the well. This is the same solution we found originally.
The other solution, x = -3, y = 2, can be interpreted similarly,
where we start by filling the 5 gallon jug (see Figure 4.3).
In a similar manner. Problem 4.3(b) gives rise to the Diophantine
equation 4x -\- 6y = 1. However, we have seen that this equation does
not have integer solutions (2 divides the left side of the equation but
does not divide 1). So, as we suspected. Problem 4.3(b) has no solution.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
119
Now try Problem 4.4 (page 101) again, if you did not solve it
the first time.
An algebraic attack may also be used for Problem 4.4. If x represents
the number of dollars andy represents the number of cents on the check,
then the check was in the amount of x dollars and y cents or,
equivalently, lOOx -f y cents. By mistake, the teller gave out y dollars
and X cents or lOOjy -f x cents. He later gave 111 cents more, and then
doubled the total disbursement to reach the correct figure. We thus
obtain the following equation:
2(1003; -f X -f 111) = IOOjc +3;.
Collecting terms, this becomes
98x - 1993; = 222.
This is a Diophantine equation, since x must represent a whole number
of dollars and y must represent a whole number of cents. In fact,
from the conditions of the problem, it is clear that both x and 3; must be
greater than 0 or the interchange could not take place. Moreover,
y < 99, since checks are not written with larger quantities of cents. By
the same token, even a badly confused teller would have been forced to
reexamine the situation if he found himself counting out a dollar's worth
of cents or more; so x must be < 99. Hence, we may assume that
0 < X < 99 and 0 < 3; < 99.
Since 98 and 199 (the coefficients of x and y in the preceding
equation) are relatively large numbers, it is not so easy to solve this
Diophantine equation by trial and error. We need more powerful
mathematical techniques. (We know that solutions exist since
gcd(98, 199) = 1, and 1|222.)
Solution of
Problem 4.4
In general, there are several techniques for solving linear Diophantine
equations. The most common makes use of the Euclidean algorithm.
We do not present this method here. (You can find a discussion of the
method in any book on elementary number theory—see, for example,
[47], p. 7). Instead, we present a method that uses the notion of
congruence (which will soon be discussed). This method often requires
a little ingenuity but, if used thoughtfully, it is generally a faster way of
solving most recreational problems of this type.
DIVISION WITH REMAINDERS
Given two natural numbers n and d, we may divide n by d. If d
does not divide n (evenly), then there will be a remainder r, which

120
CHAPTER 4
iKcernvkoiK;^
will be less than d (otherwise we would continue dividing).
For example,
3rr
6
T
2 r\
3yT-
7 = 3-2+1.
In this example, 1 is called the remainder and 2 is called the
quotient (3 is the divisor and 7 is the dividend).
In general, we have the following result:
Theorem 4.7 (Division Algorithm) Given any two positive
integers n and d, there exist integers q and r (respectively called the
quotient and the remainder) such that
n = qd -\- r and 0 < r < d.
Theorem 4.7 can be generalized for all integers n. For example, if
- 5 and d = 3, then
-5 = 3(-2) H- 1.
Note that r = 1 is positive.
Similarly, if n = -31 and d = 5, then
-31 = 5(-7) + 4.
Again, r = 4 is positive.
PRACTICE
PROBLEMS
4.H
C^v^
1. [a] Find the quotient and remainder when 108 is divided by 3.
2. Find the quotient and remainder when 129 is divided by 7.
3. [a] Find q and r such that 87 = 7^ -f r, where 0 < r < 7.
4. Find q and r such that 91 = 11^ -f r, where 0 < r < 11.
5. [a] Find q and r such that - 93 = 5^ -f r, where 0 < r < 5.
6. Find q and r such that -41 = 8^ -f r, where 0 < r < 8.
CONGRUENCE
Given a fixed divisor d, we can classify integers according to the
remainder r obtained when the integer is divided by d. We can talk
about a remainder class modulo </, consisting of all integers having
the same remainder when divided by d. Since there are d possible
remainders, 0, 1, 2, ..., d— 1, there are d remainder classes. For
example, if d = 3, then we get three remainder classes:

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
121
0= {..., -9, -6, -3,0,3,6,9,...}
T= {..., -8, -5, -2, 1,4,7, 10,...}
2 = {..., -7, -4, -1,2,5,8, 11,...}
All numbers in remainder class 0 have remainder 0 when they are
divided by 3. They are all divisible by 3 and can be expressed in the
form ?>k. _
For example, 231 is an element of 0, because
231 = 3 • 77 -f 0 = 3 • 77.
Similarly, — 9 is an element of 0:
- 9 = 3( - 3) -f 0 = 3( - 3).
All numbers in the remainder class 1 have remainder 1 when they
are divided by 3. They are of the form ?>k + \. For example, 241 and
- 23 are elements of T, because
241 = 3 • 80 -f 1
and
-23 = 3(-8) + 1.
In a similar manner, the remainder class 2 contains all numbers
that have remainder 2 when they are divided by 3. These numbers are
of the form 3/j -f 2.
Note that, once one element of a remainder class modulo 3 is known,
other elements of the class may be obtained by adding 3's to or
subtracting 3's from the known number.
1. [a] Find the remainder classes when if = 4.
2. Find the remainder classes when d = 5.
3. [a] How many remainder classes are there when d
12? 35? n?
PRACTICE
PROBLEMS
4.1 (pi^:n'^
Definition 4.8 If a and b have the same remainder when divided by
d, then we say that a is congruent to b modulo d, symbolically
denoted by a = b (mod d).
In other words, a = b (mod d) if and only if a and b are in the
same remainder class modulo d. For example.
as 241,
241 = -23 = 1 (mod 3),
23, and 1 are all in the remainder class I.

122
CHAPTER 4
The concept of congruence is a powerful mathematical tool first
formulated by Carl Friedrich Gauss in the 1790s.
From Definition 4.8, it is clear that
a = b (mod d) if and only if b = a (mod d).
Also, for any number a,
a = a (mod d).
Finally,
if a = b (mod d) and b = c (mod d), then a = c (mod ^.
Note also that
X = r (mod ^ if and only if x = kd -\- r,
for some integer k.
In addition:
If a = 6 (mod J), then a and b
belong to the same remainder
class, say, r. But then
a = qd + r and b = kd -\- r,
so
b - a = (kd -^ r) - (qd + r)
= (^ - ^)^;
thus
d\(b-a).
Conversely, one can show that
if d\(b - a) then a =b (mod d).
In other words, a = b (mod if) if
and only if d\(b - a). This fact
is often used as the definition of
congruence.
241 = -23 (mod 3)
241 and -23 are both in
remainder class 1
241 = 80 • 3 + 1
and
-23 = (-8)3 + 1
241 - (-23) = (80 • 3 + 1)
-[(-8)3+ 1]
= [80-(-8)]3
= 88-3
31 [241 -(-23)],
that is, 31264.
In the other direction, since
29 - 17 = 12
3|12
therefore
29 = 17 (mod 3)
(note that both leave remainders
of 2 modulo 3).
PRACTICE
PROBLEMS
4J
1. [a] Which of the following are true ?
(a) 7 = 1 (mod 3) (b) 7 = -2 (mod 3)
(d) 7 = 7 (mod 3) (e) 7 = 147 (mod 3)
(c) 7=3 (mod 3)
(0 76 = -152 (mod 3).

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
123
2. Which of the following are true?
(a) 8 = 3 (mod 5) (b) 8 = - 3 (mod 5) (c) 8 = - 2 (mod 5)
(d) 8 = 193 (mod 5) (e) 721 = -484 (mod 5).
3. [a] Find the smallest nonnegative integer to which 122 is congruent
modulo 3.
4. Find the smallest nonnegative integer to which 122 is congruent
modulo 7.
5. [a] Find a number divisible by 5 to which 2 is congruent modulo 6.
6. Find a number divisible by 7 to which 2 is congruent modulo 9.
7. [a] If jc = 1 (mod 3), find an expression for x in the form x =
8. If jc = 3 (mod 5), find an expression for x in the form x = .
9. Prove that a = b (mod d) if and only if d\(b - a).
r.
The following theorem presents some important properties of
congruence.
Theorem 4,8 U a = b (mod d) and e =f (mod d), then
(i) a -^ e = b -^f (mod d)
(ii) a - e = b - f (mod d)
(iii) ae = bf (mod d).
We can add, subtract, and multiply congruences with the same
modulus. In particular, since c = c (mod J), for any integer c, we
obtain
(iv) a ± c =b ± c (mod d)
(v) ac = be (mod d).
As a consequence of Theorem 4.8, in any congruence modulo d, any
number may be replaced by any other number in the same congruence class
modulo dy and the congruence will be preserved.
For example,
if 207jc = 4 (mod 5), then 2jc = 4 (mod 5) since 207 = 2 (mod 5).
That is, we can replace 207 by 2, since they are congruent modulo 5.
As another example, suppose we wish to find the remainder when
3^^ is divided by 7. This could be done by first evaluating V^ and
then dividing by 7. This would involve a lot of computation. A simpler
approach is to express the problem in terms of congruence:

124
CHAPTER 4
Find X such that V^ = x (mod 7).
Since 3-3 = 9=2 (mod 7),
X = 3^^ = (3 • 3)^^ = 2^^ (mod 7).
Since 2 • 2 • 2 = 8 = 1 (mod 7), we write 2^' as (2 • 2 • 2/ • 2. Thus
X = 2^-^ = (2 • 2 • 2)^* • 2 = I'* • 2 = 2 (mod 7).
Therefore, when 3^^ is divided by 7, the remainder is 2; that is,
y^ = 2 (mod 7). We could also arrive at this result by observing that
Then
3^ = 27 = - 1 (mod 7).
x = y^ = {Vf • 3^ = (- 1)« • 9 = 9 = 2 (mod 7).
PRACTICE
PROBLEMS
4.K
1. [a] If jc = 5 (mod 8) and j' = 6 (mod 8), to what is jc H- jy congruent
modulo 8? To what is xy congruent modulo 8?
2. If X = - 2 (mod 7) and j' = 3 (mod 7), to what is jc + jy congruent
modulo 7? To what is xy congruent modulo 7?
3. [a] Simplify each of the following congruences by reducing the
coefficients:
(a) 214x = - 16 (mod 5) (b) 26x = 413 (mod 3).
4. Simplify each of the following congruences by reducing the
coefficients:
(a) 39x = 84 (mod 7) (b) - 13jc = 25 (mod 3).
5. [a] Transform each of the following equations to congruences
modulo 7:
(a) Ix + 153; = 1 (b) -37jc + 5I3; = 212.
6. Transform each of the following equations to congruences modulo
13:
(a) 42jc + 273; = 182 (b) 37jc - 263; = 69.
7. [a] What is the remainder when 14^° is divided by 11?
8. What is the remainder when 5^^ is divided by 3?
9. [a] If 8^^ = JC (mod 5) and if 0 < jc < 4, find x.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
125
Another important theorem concerning congruences is the following:
Theorem 4.9 If a = b (mod d) and if c\d, then a = b (mod c). That
is, if d\(a - b) and c\d^ then c\{a - b).
For example, 22 = 7 (mod 15), so 22 = 7 (mod 5) since 5| 15, and
22 = 7 (mod 3) since 3115. Similarly, if x = 13 (mod 15), then
X = 13 = 1 (mod 3) and X = 13 = 3 (mod 5).
CASTING OUT NINES
The theory of congruences provides the explanation of a very interesting
phenomenon.
Definition 4.9 The digital root of a natural number is the one-digit
number obtained by adding all the digits of the original number to
obtain a new number, then adding all the digits of the new number to
obtain a third number, and so on until a one-digit result is obtained.
(Observe that the sum of the digits of a number is less than the number,
unless the number has only one digit, so that, by continuing the process
above, a one-digit number must eventually be obtained.)
For example, the digital root of 2965184860 is 4:
2 + 9 + 6 + 5+1+8 + 4 + 8 + 6 + 0 = 49
4 + 9 = 13
1 + 3 = 4.
1. [a] Find the digital roots of the following numbers:
(a) 7349621 (b) 100000 (c) 4716318
2. Find the digital roots of the following numbers:
(a) 8940361 (b) 180000 (c) 182736549
PRACTICE
PROBLEMS
41 rm:n^::i
The interesting phenomenon mentioned above is that the remainder
obtained when any natural number is divided by 9 is just the digital
root of that number (unless the digital root is 9, in which case the
remainder will be 0). In other words, any number is congruent modulo
9 to its digital root.
The justification of this phenomenon depends on Theorem 4.8 and
on the fact that our number system uses the base ten (see Chapter 5).
Specifically,
10 = 1 (mod 9)
(10 = 1 -9 + 1),

126
CHAPTER 4
hence, by Theorem 4.8,
10^ = P = 1 (mod 9)
10^ = P = 1 (mod 9)
and so on,
10" = r = 1 (mod 9)
(100 =11-9+1)
(1000 =111-9+1)
(10"= 11 ... 1 -9 + 1).
Any number may be expressed as a sum of multiples of powers of
ten. For example:
The number written as
ABCDEF in our number system
may be expressed as
ABCDEF =
A - 10^ + B • 10-* + C - 10^ +}(4.4)
D • 10^ + E • 10 + F • 1
(Recall, 10«= 1.)
By Theorem 4.8, this is
congruent modulo 9 to
1 + B • 1 + C 1
+ D-1+E-1+F-1
= A + B + C + D + E + F.
IfA + B + C + D + E + Fis
not a one-digit number, we
continue the process until a
one-digit number is obtained.
= 2-l+71+51+31
+ 1-1+4-1 (mod 9)
=2+7+5+3+1+4
= 22
= 2 • 10 + 2
= 2 + 2 (mod 9)
= 4
Thus
275314 =4 (mod 9).
The general proof follows the same lines.
The fact that we are interested only in the result modulo 9 means
that we can work step-by-step modulo 9 when we compute the digital
root. That is, we may throw away multiples of 9. Hence the term
"casting out nines.'' For example, in computing the digital root of
2965184860, we may throw away the 9, the 1 and the 8 (which add to 9),
and the 4 and the 5, leaving 2 + 6 + 8 + 6 = 22 ;2 + 2 = 4.
The process of casting out nines provides a quick check (not
foolproof) of addition and multiplication problems. If d(m) and d(n)
275314
= 200000 + 70000 + 5000
+ 300+10 + 4
= 2 - lO'^ + 7 • 10^ + 5 - 10'
+ 3 • 10^ + 1 - 10 + 4 • 1

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
127
denote the digital roots of w and n respectively, then m = d(m) (mod 9)
and n = d(n) (mod 9), and so, by Theorem 4.8,
and
d(m -\- n) =m + n = d(m) -f d(n) (mod 9)
d(mn) =mn = d(m)d{n) (mod 9).
Thus, the digital root of a sum (product) of numbers should be the
same as the digital root of the sum (product) of the digital roots of the
numbers.
For example, is the following calculation correct?
269 X 347 = 93543
The digital root of 269 is 8, that of 347 is 5, and that of 93543 is 6.
Since 8 x 5 = 40, which has digital root 4 rather than 6, there is a
mistake in the multiplication.
The process of casting out nines is also helpful in designing some
mystifying number guessing tricks. (See Exercises 4.25 through 4.28
at the end of the chapter.)
1. [a] Find the missing digit if 3_4103161 is divisible by 9.
2. Find the missing digit if 4731_29 leaves a remainder of 2 upon
division by 9.
3. [a] What is the remainder when 456823940 is divided by 9?
4. Use digital roots to find the mistake in the following:
3641
X 128
29128
7282
3641
468048
N 5. [s] Use equation 4.4 to find a rule that gives the remainder when
ABCDEF is divided by 2; by 3; by 4; by 5; by 7; by 8; by 11.
6. What is the remainder when 24732151 is divided by 2? by 3? by 4?
by 5? by 7? by 8? by 11?
PRACTICE
PROBLEMS
4.M
SOLVING LINEAR CONGRUENCES
Let us return to our goal of developing machinery for solving linear
Diophantine equations.

128 CHAPTER 4
The properties of congruence exhibited in Theorem 4.8 are almost
identical to the properties of equality. However, there is one property
of our number system that holds for equality but does not hold for
congruence—namely, the cancellation law:
If ax = ay^ and a # 0, then x = y.
An analog for congruence,
if ax = ay (mod d) and a ^0 (mod d), then x =y (mod d)
does not always hold. For example, putting x = 3, jy = 7, a = 6, and
^ = 8, we see
6-3= 18 =42 = 6-7 (mod 8).
But 3^7 (mod 8), so we cannot cancel the 6.
There is a case, though, when the cancellation law for congruence
multiplication does hold:
Theorem 4.10 If ax = ay (mod d) and if gcd(a, ^ = 1, then
X =y (mod d).
Note the condition that gcd(a, d) = 1; in this case we can cancel.
For example, if 3x = 15 (mod 34), then since gcd(3, 34) = 1,
jc = 5 (mod 34).
Actually, we can cancel in a more general setting, but we must
change the modulus. That is, we have the following theorem:
Theorem 4.11 If ax = ay (mod d) and if gcd(a, d) = g, then
X =y (mod dig).
(Note that Theorem 4.10 is a special case of Theorem 4.11, that is,
g = gcd(a, d)= 1.)
For example, if
6x = 6y (mod 8),
then
x=y (mod 4) (4 = 8/2 = 8/gcd(6, 8)).
Thus,
6 • 3 = 6 • 7 (mod 8) implies 3=7 (mod 4).
Similarly, if
8jc = 12 (mod 20),
then
2jc = 3 (mod 5) (5 = 20/4 = 20/gcd(8, 20)).

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
129
(a) (a) If Ix = 35 (mod 81), solve for x.
(b) If Ix = 35 (mod 84), solve for x.
Solve each of the following congruences for x:
(a) 3x = 27 (mod 42). (b) -8x = 16 (mod 23).
(c) 4x = 24 (mod 30).
PRACTICE
PROBLEMS
4.N
Another important difference between equality and congruence is
that we can always solve the linear equation ax = 6, as long as a # 0
(just let X = bja); but we cannot always solve linear congruences. For
example, the congruence 2x = 3 (mod 4) has no solutions. (When we
deal with congruences, we are only interested in integers. For any
integer x, 2x will be even and will leave a remainder of 0 or 2 when
divided by 4; it will not leave a remainder of 3.)
In general, given the congruence ax = b (mod d), if x = n is a
solution, then any integer congruent to n modulo d is also a solution,
since we have seen that any number in a congruence may be replaced
by any other number in the same remainder class.
When we count the number of solutions of a congruence, we mean the
number of distinct remainder classes in which solutions lie.
For example, x = ?> satisfies 2x = 1 (mod 5); hence, x = 8 also
satisfies it. 2(8) =16 = 1 (mod 5). Similarly, x = ..., - 7, - 2, 3, 8, ...
all satisfy this congruence. We write x = 3 (mod 5) is a solution.
One way of determining the solutions of a congruence modulo d
is to substitute 0, 1, 2, ..., if - 1 for x, testing when the congruence
is satisfied. Thus, as we saw above, 2x = 3 (mod 4) has no solutions
since 0, 1, 2, and 3 do not satisfy this congruence.
Now consider 3x = 1 (mod 5). Substituting
x = 0, 30 = 0^1 (mod 5)
x= 1, 3 1=3^1 (mod 5)
x = 2, 3-2 = 6 = 1 (mod 5)
x = 3, 3-3 = 9^1 (mod 5)
X = 4, 3 • 4 = 12 ^ 1 (mod 5),
we see that this congruence has a unique solution modulo 5: x = 2
(mod 5). Note that x = 2, x = 7, x = 12, ... all satisfy the congruence.
As a further example, consider 2x = 0 (mod 4). Substituting
x = 0, 20 = 0=0 (mod 4)
x=l, 21=2^0 (mod 4)

130
CHAPTER 4
jc = 2, 2-2 = 4=0 (mod 4)
x = 3, 2-3 = 6^0 (mod 4),
we see that this congruence has two solutions modulo 4: x = 0 (mod 4),
and X = 2 (mod 4).
The general situation is given in the following theorem:
Theorem 4.12 The congruence ax = b (mod d) is solvable if and
only if gcd(a5 d) divides b. If it is solvable, then there are^ = gcdia, d)
solutions modulo d.
Note that, if gcd(a, d) = 1, then, according to Theorem 4.12,
ax = b (mod d) has a unique solution modulo d^ since 1 certainly
divides b.
PRACTICE
PROBLEMS
4.0
[a] How many solutions are there to each of the following
congruences? That is, how many remainder classes give solutions?
(a) 3x = 6 (mod 8) (b) 3x = 6 (mod 9)
(c) 3x = 1 (mod 6) (d) 4x = 26 (mod 98).
How many solutions are there to each of the following congruences ?
That is, how many remainder classes give solutions?
(a) 4x = 24 (mod 48) (b) 3x = 1 (mod 17)
(c) 9x = 3 (mod 6) (d) 6x = 27 (mod 45).
[a] Which of the following congruences can be solved? Solve, by
trial, if possible:
(a) 5x = 7 (mod 12) (b) 3x = 7 (mod 12)
(c) 12x = 8 (mod 16) (d) 5x = 8 (mod 3).
Which of the following congruences can be solved? Solve, by trial,
if possible:
(a) 5x = 9 (mod 12) (b) 3x = 9 (mod 12) (c) 3x = 11 (mod 18).
The trial and error method of solving linear congruences works
well if the modulus is small, but it is not practical for larger moduli.
Another method of solving congruences is as follows: We have
3x = 1 (mod 5). We wish to replace 1 by a number that is congruent
to it modulo 5 and that is divisible by 3. In this case, 6 works. (So
does - 9.) Thus 3x = 1 (mod 5) has the same solutions as 3x = 6 (mod 5).
Since gcd(3, 5) = 1, we can cancel 3 by Theorem 4.10. Thus x = 2
(mod 5). [If we had used — 9 instead of 6, we would have obtained
X = — 3 (mod 5), which is the same solution since -3=2 (mod 5).]

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
131
In general, if we have
ax = b (mod d) with
gcd(a, d) = 1 and if a \ b, then
we wish to find a number that is
divisible by a and is in the same
remainder class as b. To do this
we keep adding (or subtracting)
d to b until we reach a number
that is divisible by a.
so
7x = 6 (mod 37)
6 + 37 = 43 (7 I 43)
43 + 37 = 80 (7 I 80)
80 + 37 = 117 (7 I 117)
117 + 37 = 154 (71154)
7x = 154 (mod 37)
jc = 22 (mod 37)
[Note 154 =6 (mod 37).]
1. [a] Solve each of the following:
(a) 5jc = 3 (mod 12) (b) 5jc = 7 (mod 12)
(c) 15jc = 39 (mod 59).
2. Solve each of the following:
(a) 7jc = 11 (mod 16) (b) 4jc = - 39 (mod 11)
(c) 4jc = -38 (mod 41).
PRACTICE
PROBLEMS
4.P
(Tsmm^
SOLVING LINEAR DIOPHANTINE EQUATIONS
We are now ready to apply what we have learned about congruences
to solve linear Diophantine equations of the type ax + by = c. As we
saw above, there can be no solution unless gcd(a, b)\c.
If gcd(<2, b) does divide c, then
we can divide through by
gcd(<2, b) to obtain
a^x + b'^y = c*
where
algcd{a, b),
blgcd(a, b),
clgcd(a, b).
This new equation has exactly
the same solutions as the original
equation. Furthermore, since we
66x + 483; = 30
gcd(66, 48) = 6; 6|30
llx + 83; - 5

132
CHAPTER 4
have divided out the greatest
common divisor of a and 6, the
resulting numbers, a* and 6*,
can have no common divisor
greater than 1. That is,
gcd(a*, b*)= 1.
gcd(ll,8)= 1.
Therefore, there is no loss in generality if we consider only
Diophantine equations ax + by = c for which gcd(a, b) = \.
We continue with the general method for solving ax + by = c on
the left side of the page, and with the example 1 Ijc + 8jv = 5 on the
right.
Consider the new equation as a
congruence modulo either a or b.
(Usually, we choose the smaller
of the two so that we deal with
smaller numbers.) Working
modulo 6, we get
ax = c (mod b).
Since gcd(a, 6) = 1, this
congruence has a solution. If
possible, we reduce a and c
modulo b to obtain
a'x = c' (mod b).
As indicated above, we can find
the solution of this congruence
by adding or subtracting
multiples of b to c', until we
obtain c" which is divisible by
a'. Then, the solution is
X = d (mod /?),
where d = c"la'. Or,
equivalently,
X = d -^ kb,
where k = 0, ±1, ±2, ... .
For each value of /?, we may
substitute the corresponding
lljc + 83; = 5
lljc + 83; = 5 (mod 8)
Since 8=0 (mod 8),
lljc -f O3; =5 (mod 8)
llx = 5 (mod 8)
11=3 (mod 8), so
3x = 5 (mod 8)
3jc = 5-8 = -3 (mod 8)
3x = - 3 (mod 8)
jc = - 1 (mod 8)
(where - 1 = - 3/3)
X = - 1 + 8/j,
k = 0, ±1, ±2, ...

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
133
value of X into ax + by = c to
find the corresponding value for
y-
Usually, the conditions of the
problem limit the values that
need to be considered for Xy and
hence limit the values for k that
must be tried.
Sy = 5 - Wx
= 5 - 11(-1 + 8/^)
= 5+11-88/2
= 16 - 88/^
y = 2- \\k
The solution is
x= -\ +Sk, y = 2 - Ilk,
k = 0, ±1, ±2, ... .
If the problem restricts jc to be a
number between 10 and 20, k
must = 2, X = 15, y = - 20.
1. [a] Find all solutions for \lx - I9y = 1, for which 50 < jc < 100.
2. Find all solutions for 11 x - I9y = 3.
3. [a] Find all solutions for 31 x - 53y =11.
4. Find all solutions for 44x H- 233; = 13.
PRACTICE
PROBLEMS
We are now ready to consider Problem 4.4 (page 101).
If you have not already solved it, try it again now.
Earlier we obtained the equation 98x — \99y = 222, where jc represents
the number of dollars and y represents the number of cents on the
check. Observe that gcd(98, 199) = 1. Since 98 < 199, we consider the
equation as a congruence modulo 98:
98jc - 1993; = 222 (mod 98).
Since 98 = 0 (mod 98), 199 = 3 (mod 98), and 222
reduces (by Theorem 4.8) to
- 33; = 26 (mod 98).
26 (mod 98), this
Since 26 is not divisible by 3, we add 98, obtaining 124. [This doesn't
change the congruence because 124 =26 (mod 98).] As 124 is still not
divisible by 3, we add 98 again, obtaining 222. Thus
Solution Of
Problem 4.4
(Continued)
33) = 222 (mod 98)

134 CHAPTER 4
-y = 74 (mod 98)
or
y= -74 (mod 98)
or
y = - 74 + 98/j.
Since we need 0 < jy < 100, we need only consider one value of k, k = 1.
That is,
0 < -14 + 98k < 100
74 <98/^ < 174
0 < 74/98 <k< 174/98 < 2
since k is an integer. Thus, y = - 74 + 98 • 1 = 24. Substituting this
value of 3; into 98jc - 1993; = 222, we obtain 98jc = 222 + 199 • 24 =
222 + 4776 = 4998;
X = 51.
Thus, the amount of the check was $51.24. (Of course, you should
return to the original problem to check that this is the correct answer.)
There are times when a congruence is not solved so easily. For
example, consider the Diophantine equation
285jc + 39I3; = 72.
Transforming this to a congruence modulo 285,
285jc + 39I3; = 72 (mod 285).
Reducing modulo 285,
Ojc + IO63; = 72 (mod 285)
IO63; = 72 (mod 285).
Since gcd(2, 285) = 1, we can divide both sides of the congruence by 2:
53y = 36 (mod 285). (4.5)
If we now proceed as previously, we would start adding (or
subtracting) 285 to 36 until we obtain a number that is divisible by 53:
36 + 285 = 321; 53 \ 321
321 + 285 - 606; 53 \ 606
etc.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY 135
Although we know that we must eventually reach a number divisible
by 53 (this is guaranteed by the fact that gcd(53, 285) = 1), it might
take a long time for this to happen. Furthermore, deciding whether
or not a number is divisible by 53 is not easy. It therefore would be
preferable for us to find an alternate approach.
Suppose we take the congruence 4.5 and write it as an equation:
533; = 36 + 285^, (4.6)
for some integer q. This is again a Diophantine equation, and, as
before, we can replace this by a congruence:
533; = 36 -f 285^ (mod 53) (4.7)
0 = 36 -f 285^ (mod 53)
0 = 36 -f 20q (mod 53)
-20q = 36 (mod 53). (4.8)
The important advantage of congruence 4.8 over 4.5 is that the modulus
is smaller (53 as opposed to 285) and it should be easier to solve.
Dividing by 4, (note gcd(4, 53) = 1),
-5^ = 9 (mod 53) (4.9)
we could now proceed by trying to add (or subtract) 53 to 9 until
we obtain a number divisible by 5—since 5 is a smaller number, this
would be a reasonable approach—or we could again replace congruence
4.9 by an equation
-5^ = 9-f53f (4.10)
for some integer t. This equation can again be replaced by a
congruence :
- 5^ = 9 -f 53r (mod 5)
0 = 4 -f 3r (mod 5)
-3f =4 (mod 5). (4.11)
Observe that the modulus (5) of this congruence is smaller than
previous moduli. In fact, 5 is small enough that congruence 4.11 can
be easily solved by trial and error. The solution is
t = 2 (mod 5).
We now wish to use this to solve congruence 4.5, so we can find y.
Since knowing one solution of congruence 4.5 determines all solutions,
we can take r = 2 rather than having to deal with t = 2 (mod 5).
Substituting t = 2 into equation 4.10,
- 5^ = 9 -f (53)(2) = 9 -f 106 = 115
q= -23.

136 CHAPTER 4
Substituting ^ = — 23 into equation 4.6,
5?>y = 36 + 285(-23) = 36 - 6555 = -6519
y = -123.
Therefore the solution of congruence 4.5 is
y = - 123 (mod 285)
or, equivalently,
y = - 123 -f 285/j.
Substituting into the Diophantine equation and solving for x, we
obtain
285x -f 391( - 123 + 285/j) = 72
285x + (285)(391/j) = 72 + (391)(123)
285(x + 39U) = 72 + 48093 = 48165
(X -f 39U) = 169
X = 169 - 391/^.
Thus, the solution of the Diophantine equation is x = 169 - 391/j,
jV = - 123 H- 285/j. For each value of k, we get a pair of values for
X and J'.
In general, if we wish to solve
ax = b (mod m),
with gcd(a, w) = 1 and 0 < a < w, we can replace the congruence by
the Diophantine equation
ax H- my = 6,
which we in turn replace by the congruence
my = b (mod a).
Since this congruence has a smaller modulus, it may be easier to solve
by trial and error; if not, we can continue the procedure. As the
modulus keeps getting smaller and smaller, the method must yield a
solution in a finite number of steps. However, the trial method may be
used at any point along the way to speed up the procedure.
We have presented two approaches to solving a congruence of the
form ax = b (mod m). One is to keep adding (or subtracting) m to b
until we obtain a number divisible by a, and the other is to replace
the congruence by a Diophantine equation which is in turn replaced by
a congruence, etc. Which of these two approaches is preferable? That
will vary from problem to problem. If the coefficient a is reasonably

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
137
small, we suggest trying the addition/subtraction method first; but if
adding m to 6 one or two times and subtracting m from b one or two
times doesn't result in a number divisible by a, then try the other
approach. If the coefficient a is fairly large—say, greater than 20—then
we suggest using the second approach immediately.
THE CHAPTER IN RETROSPECT
In this chapter you have met some more types of problems that have
been of interest to recreational mathematicians for centuries. But this
chapter has also given you a taste of some of the basic mathematical
concepts in the field of number theory—a field that has been called
the " Queen of Mathematics." We have only touched upon the subject;
but many of the topics we have considered (divisibility, prime
factorization, congruence) have analogs that are also important in other
branches of mathematics.
This chapter is different from the first three in that it has formally
presented definitions and theorems. Precise definitions are important in
mathematics. We need them so that, when we refer to a particular
term, we all know exactly what we mean and we all mean the same
thing. For example, when we say that a number is prime, we should
all understand that we are really saying that the number has exactly
two divisors; we should also understand that we will be able to
determine whether or not a number is prime if we can count its
divisors.
Theorems are also important. They elaborate on the properties of
and the relationships between the terms that are defined. They give us
general information about all mathematical objects of some particular
type, rather than just telling us about a particular example. For
example, Theorem 4.12 gives us information about all congruences of
the form ax = b (mod d). It tells us when such a congruence has a
solution. The theorem is a general result. It tells us much more than
that a particular congruence, such as 3x = 1 (mod 5), has a solution;
it tells us that all congruences of the particular type which satisfy
certain conditions have solutions.
Theorems must be proved before they are accepted as valid. Proofs
are usually based on definitions, on the statements of other, previously
proven theorems, and on primary assumptions called axioms which
are accepted without proof.
This chapter has supplied very few actual proofs. Although most
of the theorems that were presented are not difficult to prove, we
have been more interested in discussing their applications rather than
giving a rigorous presentation.
The theorems have been used to solve the sample problems. They

138
CHAPTER 4
also will be helpful for many of the exercises that follow. In general,
problems of the type dealt with in this chapter can be stimulating;
but they also can be frustrating. It is exciting to see how a little
mathematical theory can be useful in solving such problems.
Exercises
Primes and Divisibility
4.1. A group of sorority sisters stopped for
refreshments at the Soda Shoppe. Each
ordered a Coke, and the total bill was $1.87,
exclusive of tax. They were surprised to find
out that a Coke was no longer $.15; prices
are going up. How many were in the group,
and what is the price of a Coke?
4.2. [a] a golden age group went on a bus
tour. Each person paid his or her own exact
fare with five coins. In totaling the fares, the
bus driver obtained a sum of $21.83.
How many pennies did the bus driver
receive?
4.3. [h] Using a desk calculator, a student was
asked to obtain the complete factorization of
24,949,501. Dividing by successively
increasing primes, he found the smallest prime
divisor to be 499 with quotient 49,999. At this
point, he quit.
Why didn't he continue his division? [52],
Problem 16)
4.4. [a] When the accountants for Lose-a-
digit Computer, Inc. had finished preparing
their annual budget, they presented the final
figures to the president, I. M. Smart.
'' It looks like a good year," he exclaimed.
" The amount of the budget just happens to
be the smallest number of cents (other than
one cent) that is a perfect square, a perfect
cube, and a perfect fifth power."
How much money has Lose-a-digit
budgeted for the year?
4.5. [a] a census taker stopped at Hotel
Sleep-Inn. The clerk at the desk was a
mathematics student with a sense of humor. When
asked how many guests were in the hotel, she
replied, "The number is the smallest positive
integer that has the following properties:
When divided by two, the result is a
perfect square;
when divided by three, the result is a
perfect cube."
The census taker, who had been calculating
all day, slammed his book on the table. "I
quit," he said.

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
139
Can you figure out how many guests were
in the hotel?
4.6. [h] [a] If w is a positive integer, then w!
(read "w factorial") denotes the product of all
positive integers that are less than or equal to n.
For example,
5! = 5-4-3-21
120,
6! = 6-5-4-3-21 = 720,
10! = 10 -9 • 8 • 7 • 6 • 5 -4 • 3
= 3,628,800.
2 • 1
Note that 5! and 6! both end in one zero,
and that 10! ends in two zeros. Without
actually computing 50!, can you tell the
number of zeros in which 50! ends ? (Note:
50! is not equal to 5! • 10!)
M 4.7. [s] [a] The Wellington children
wanted to buy an anniversary present for
their parents. They each contributed a number
of dollars equal to the number of boys in the
family, and thereby collected a number of
dollars that exceeded by 24 the number of years
that their parents had been married.
The number of sons, the number of
daughters, and the number of years that the
parents had been married were three distinct
prime numbers.
Find these numbers.
4.8. [h] a Christmas tree is decorated with
36 lights, which are numbered 1 through 36.
Timers are set so that every 5 minutes a
change occurs in the light pattern. The
sequence of changes repeats every 3 hours.
Switches are set so that:
At the end of the first time interval, every
light is turned on.
At the end of the second time interval,
every second switch is reversed.
At the end of the third time interval, every
third switch is reversed.
Etc.
(a) What is the state of the lights at the
end of the first 3 hours (after the 36th time
interval) ?
(b) If the tree had n lights, what is the
state after n time intervals?
^ 4.9. [a] Three boys, Pedro, Quincy, and
Ralph, and their sisters, Sandy, Trixie, and
Vera (not necessarily in that order), had
chickens for pets.
Last week was an unusual one. Each chicken
laid as many eggs as its owner owned chickens.
Quincy had three times as many chickens as
his own sister, and had eight more chickens
than Ralph's sister. Furthermore, by the end
of the week, Quincy had collected 56 more
eggs than Pedro; Ralph had collected 52 more
than Sandy; and Pedro had collected as many
eggs as Sandy and Trixie together.
How many chickens did each of the six
people own? And, who was whose sister?
¥ 4.10. [s] [a] If a' = b^ and c' = cP, where
a, b, c, and d are positive integers; and if
c - a = 25, what are a, 6, c, and d? ([60],
Vol. 3, Problem 174)
4.11. [h] Find three positive integers whose
sum is 25 and whose product is 540.
M 4.12. [h] [a] Once again, the census taker
appears on the scene. He stops at a house,
notes down the number on the door, and
knocks. When a woman answers, he asks her

140
CHAPTER 4
age and notes the answer. Then he asks if
anyone else lives at the house. She replies that
three children live with her. Upon asking their
ages, the census taker is given the reply that
the sum of their ages equals the number on the
door and that the product of their ages equals
36. The man does some quick computation and
says that he needs another clue. He then asks
if the youngest of the three is a twin. The
woman replies that he is not; whereupon, the
census taker is able to compute the ages of the
three.
What are their ages?
4.13. [h] Find the largest number that leaves
the same remainder when it is divided into
887, 959, 1007, and 1187.
Linear Diophantine Equations
4.14. The annual dues for the Burbank Book
Club are generally $23.00 per person.
However, senior citizens pay only $17.00. If the
total amount of dues collected this year was
$ 1500.00, what is the smallest number of senior
citizens that could belong to the club?
4.15. [a] Judy had absolutely no money for
shopping, so she went to the bank to cash her
tax refund check. Somehow, a $50 bill had
become mixed in with the $20 bills in the
teller's drawer, and the teller mistakenly gave
this bill to Judy, thinking that it was a $20
bill.
Judy did not notice the mistake until after
she had spent $2.40 in one store for some
pairs of stockings and was about to purchase a
dress in another store. The price of the dress
(including tax) was exactly one-third of the
money she then had with her. When she
removed the "$20" bill from her wallet, she
discovered that it was $50. She paid for the
dress anyway, and observed that this left her
with a number of cents equal to the amount of
dollars of the original check and a number of
dollars equal to the amount of cents of the
check.
Being an honest person, Judy returned to the
bank to correct the mistake.
How much was Judy's tax refund?
y/;ifr»^wjrf>^^.
4.16. A used car salesman purchased a
number of very old and battered cars during
the past month. He paid the same price, $89,
for each. He has not yet sold all of the cars,
but those that he has sold went for $225 apiece.
He has already made a profit of $2,327 (total
sale price minus outlay for all cars purchased).
If his car lot has room for no more than 50
cars at one time, how many cars remain?
4.17. [h] One hundred units of U.S. money
were worth a total of $100. They included
only half-dollars, $5 bills, and $10 bills. How
many coins and bills of each denomination
were there?
4.18. [a] At a meeting of the Coin Collectors
of America, every member present had fifty
U.S. coins (pennies, nickels, dimes, quarters,
and half-dollars) in his pockets. Each person's
coins added up to $1.00, and no two people
brought exactly the same coins.
What is the largest number of people who
could have attended the meeting?
4.19. [s] [a] Three freshmen stopped into
their college book store for supplies. Garrett
paid $6.95 for six notebooks, four pencils, and
five pens. Beth bought two notebooks, five

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
141
pencils, and three pens for a total of $2.85,
commenting that the " ten cent pen " no longer
existed.
Assuming that each notebook, each pencil,
and each pen costs an exact number of cents,
what did Wilma pay for one notebook, one
pencil, and one pen?
4.20.1 [h] [a] The new saleswoman at Clocks
Unlimited was unbelievable. During her first
day on the job, she sold at least one of each
of the three models that other salespeople
found most difficult to sell—the chartreuse
kitchen clock that sells for $17, the $31 owl-
shaped cuckoo clock that hoots instead of
cuckooing, and the two-foot high grandfather
clock that was on sale for $61. In all, the
saleswoman collected $300 (excluding tax)
from selling these models alone.
How many of each did she sell?
Congruence and Digital Roots
4.21. Locksher Mohles had finally decoded
the message. It read, " The jewels are buried
in St. Breka Cemetery under the 221st
tombstone." The writing was followed by the
picture in Figure 4.4.
->-
J
Mohles and his close friend, Dr. Swanto,
reached the cemetery. There they found only
15 tombstones in the entire cemetery, and
these were arranged in a straight line in front
of the entrance.
"What do we do now, Mohles?'' Swanto
asked.
"Obviously, from the map, we must count
the tombstones from 1 to 15 and then we must
reverse direction, counting tombstone # 14 as
16, #13 as 17 and so on, until we reach #1
which we count as 29. Then we reverse
direction again, and so on," replied Mohles.
"But that will take a long time," snapped
Swanto; "And it's spooky here."
"Don't worry. I know right now which
tombstone to approach."
Do you know too? (No fair counting up to
221.)
4.22. What is the remainder when 3^^^^^^^ is
divided by 7?
for any positive
is always divisible
M 4.23. [h] Prove that,
integer w, 3^" + ^ -f 2" + ^
by 5.
-¥ 4.24. [h] [a] Magician: "Select any
number from 1 to 12, but do not tell me what
number you have selected. Now choose a
second number from 1 to 12, but this time
tell me the number."
FIGURE 4.4
Subject: "4."
Magician: "Fine. Now, starting with 4 and
proceeding in a counterclockwise direction, I
will tap the numbers on the face of my watch,
one at a time. When I tap the 4, say to

142
CHAPTER 4
yourself the number you originally selected;
with each successive tap I make, you add 1 to
your number. Stop me when you reach 16.
I will then be pointing at the number you
selected."
As usual, the magician's prediction was
realized.
Explain why this trick works. What would
the magician have said if the second selected
number had been a number other than 4?
4.25. [h] Pick a number. Multiply it by 3.
If the result is odd, add 25; if it is even, add
34. The new result will be even; divide it by 2.
Add 11. Now multiply by 6.
Add up the digits of your answer. If the
number you obtain contains more than one
digit, add the digits again. Continue doing so
until you obtain a one-digit result.
Your answer will be 6.
How do we know?
4.26. [a] Giselda, the magician's wife, was
learning the trade fast. She took two decks
of cards from her pocket and placed them face
down on the table. She then asked a volunteer
from the audience to pick a number greater
than 11 and smaller than 20. The volunteer
did so. She then handed him one deck of cards
and asked him to deal the cards face down on
the table, placing each card on top of the
previous one and counting one for each card,
until he reached the number he selected. She
then picked up the packet of cards that the
volunteer had just dealt, and proceeded to
arrange them like the numbers on a clock. That
is, she placed the first card in the position of
the number 1, the next card in the position
of number 2, and so on, until 12 cards were
so arranged. The remainder she returned to
the deck.
The volunteer was then told to add the
digits of the number he had selected and to
note the card that was in the corresponding
position. (For example, if the number he chose
had been 15, then he would note the card in
position number 6.)
Giselda then handed the volunteer the
second deck, asking him to announce his
chosen card and to hold the deck so that the
audience could see the face of the cards.
To everyone's amazement, all the cards the
volunteer held were identical to the card he
announced.
How was Giselda able to perform this trick
without knowing in advance what number
would be selected?
4.27. [h] The magician himself insisted on
doing the next trick. He pointed to a young
woman in the audience.
"Pick a number," he said. "Double it; add
7; multiply by 5; subtract the number you
started with. Remove any nonzero digit from
the answer; and tell me the remaining digits
in any order."
"6 and 8," replied the woman.
"Then the digit you removed is a 3,"
announced the magician.
He was correct. How did he know?
4.28. [h] As another variation of the trick in
Exercise 4.27, the magician handed a piece of
paper and a pencil to another volunteer from
the audience. He then gave the following
instructions:

SOLVE IT WITH INTEGERS: SOME TOPICS FROM NUMBER THEORY
143
"Write down any number; scramble the
digits to make another number; subtract the
smaller of the two numbers from the larger;
circle any nonzero digit in the result; add the
remaining digits and tell me the sum."
" 14," announced the volunteer.
"Then the number you circled is a 4."
Again, how did the magician know?
^ 4.29. [h] [a] If the sum of the digits of
IV is S, and the sum of the digits of 5" is T,
find the sum of the digits of T.
4.30. [h] [a] Four boys were playing close to
a bubble gum machine. Suddenly, Kevin
knocked it over, the machine broke, and the
bubble gum pieces rolled into a pile on the
floor. The three older boys—Kevin, Esteban,
and Tony—decided that they would split the
gum four ways. They left Sean, the youngest,
to watch over the pile while they went to get
containers.
Kevin was the first to return. He counted
the number of gum balls and found that, if
the number were divided by four, there would
be one left over. Feeling that he was entitled
to the extra piece of gum (since, after all, it
was he who knocked over the machine in the
first place), Kevin took the extra piece plus
one-fourth of the remaining gum and left.
Esteban, the oldest of the four, was the
next to arrive. He counted the pieces of gum
and again found there was one more than
could be evenly divided among the boys. Sean
was just about to tell him that Kevin had
already taken his share, when Esteban said,
"Since Fm the oldest, I get the extra piece.
If you don't like it. Til punch you in the
nose." Sean decided to keep his mouth shut.
So Esteban took the extra piece of gum and
one-fourth of the rest and left.
When Tony arrived, the scene was
essentially the same. He, too, found one piece
of gum too many to be divided between the
four boys and decided to keep the extra piece
for himself. He therefore took one gum ball
and one-fourth of the remainder and left.
Sean then gathered up the remaining pieces
of gum and went home.
What was the smallest number of gumballs
that could have been in the machine ? Who got
the best of the deal? Who got the worst?
4.31. [^ [a] A farmer had eight baskets, some
containing apples and the remainder
containing plums. The baskets contained 67, 62, 35,
34, 30, 25, 19, and 17 fruits respectively.
Each apple cost three times as much as each
plum. The farmer's first customer purchased
a number of baskets of apples and paid the
farmer $11.88. The second customer
purchased a number of baskets of plums, also
paying $11.88.
This left the farmer with one basket of
apples.
What was the remaining basket worth?
¥ 4.32. [h] According to the terms of the last
will and testament of J. P. Moneybags, his
entire fortune (which turned out to be worth
$2,518,000) was to be divided equally (in exact
multiples of $1000) among all of his direct
descendants who were present at the
conclusion of his funeral. The excess, if any, was
to be donated to the Moneybags Home for
Retired Mathematicians.
Mendel Giant, the director of the home,
attended the funeral to determine how well the
institution would make out.
Ten minutes before the funeral was
scheduled to begin, only three heirs were
present.
"I guess that means we'll get $1000," said
Giant.
Just then, another heir entered.
"Good," exlaimed Giant. "That means
$2000 for us."
One minute later, a fifth heir entered.
"$3000," Giant mumbled.

144
CHAPTER 4
As the funeral was about to begin, three
more heirs came in.
"Terrific," uttered Giant. "Now we'll get
$6000.''
During the funeral, three more heirs
entered, one at a time. With each one, Giant
mentioned a new figure. At the entrance of the
9th heir, Mendel said," $7000 "; the tenth heir,
"$8000"; and the final heir, "$10,000."
How was Dr. Giant able to determine so
quickly how much the home would receive?
(That is, how could he tell the remainder when
2,518 was divided by each of the numbers
above?)
^ 4.33. [h] (a) [a] For each value of w, from
2 to 16, find the largest and smallest nine-
digit numbers, containing each of the nine
nonzero digits exactly once, that are exactly
divisible by n.
(b) Do the same for ten-digit numbers
containing each of the ten digits (including
0) exactly once. (Note that 0 may not be
the first digit of a number.)
4.34. [h] Two digits of the nine-digit number
25_37_411
are missing. Find these digits if the number
is divisible by 99.
4.35. [h] [a] In how many different ways can
you fill in the blanks of
5_3_1_672
with the digits 4, 9 and 8, in some order, so
that the resulting number is divisible by 792 ?
M 4.36. The Gregorian calendar, which
most nations use today, is based on a year
containing 365 days. Every fourth year is a
leap year (containing 366 days), except that
years divisible by 100 are not leap years unless
they are also divisible by 400, in which case
they are leap years. Thus, for example, 1900
was not a leap year, but 2000 will be.
(a) [s] Show that every 400 years the
calendar repeats itself (that is, the same dates
fall on the same exact day of the week). Thus,
for example, since January 1, 1976 was a
Thursday, January 1, 2376 will also be a
Thursday.
(b) Show that the calendar also repeats itself
every 28 years that do not include the turn
of a century.
(c) [s] [a] What day of the week was
January 1, 1901?
(d) [h] [a] In what years of the 20th
century does February have five Sundays?
(e) [h] [a] What day of the week, if any,
can never be February 29?
(0 [h] [a] What day of the week, if any,
can never be the first day of a new
century ?
(g) [h] [a] Which years of the 20th century
contain 53 Sundays?
MM (h) [a] How many times in the 20th
century will the thirteenth of a month fall
on a Friday?
(Note the 20th century runs from January 1,
1901 through December 31, 2000, inclusive.)
([60], Vol. 4, pp. 6, 34; [1], Vol. 40, p. 607.
See also [47], p. 204)

More About Numbers:
Bases and
Cryptarithmetic
You have probably had an experience similar to the following: You
are given a large pile of pennies to count. You begin counting. As
you count each penny, it is taken from the original pile and placed in
a "discard'* pile. You reach, say, 1324 and you are interrupted. (A
streaker enters the room, dances a jig, and exits down a fire escape.)
You attempt to resume counting, but can no longer remember what
number you were up to. Oh well, start counting all over again.
A little foresight could have saved you a lot of wasted effort.
Suppose that, instead of placing all pennies that have been counted
in the same discard pile, you placed them in piles of 10, combining
ten of these piles into a pile of 100 when possible, and so on. When
you were interrupted at 1324, you would have had before you an
arrangement as in Figure 5.1.
100
100
M
^^
g
^^
&
k=J
g
^
K^
^3
g
^3
y
3
"^^
g.
N
^3
100
© © ©
1 1 1
©
1
145

146
CHAPTER 5
After the interruption, you would have had very little trouble
resuming your count where you had left off.
This is the basic idea behind our decimal, or base ten, notation for
numbers.
1324 = 1 • 1000 -f 3 • 100 -f 2 • 10 -f 4 • 1.
In the decimal system, we express numbers in terms of powers often;
hence the term "base ten." Historically, not all nations have used the
base ten for their number systems. The ancient Babylonians used the
base 60, and there is evidence that the bases 2, 3, 4, 5, 12, and 20
have also been used ([14], pp. 8-9).
In this chapter, we consider arithmetic in base ten and other bases.
We also consider problems in which we are given only partial
information about some arithmetical calculation. Some typical recreations of
this type follow. Try them before reading further.
SAMPLE Problem 5.1
PROBLEMS
An assayer owns a balance scale that he uses to weigh ore samples.
He always places the samples on the left pan of the balance and
places weights on the right pan until a balance is achieved. If he
wishes to be able to balance all possible samples weighing an integral
number of grams between 1 and 127, what is the smallest number of
weights he will need?
Problem 5.2
When the first Martian to visit Earth attended a high school algebra
class, he watched the teacher show that the only solution of the equation
5x^ - 50x -f 125 = 0 is X = 5.
" How strange," thought the Martian. " On Mars, x = 5 is a solution
of this equation, but there is also another solution."
If Martians have more fingers than humans have, how many fingers
do Martians have?* ([38], Chapter 7, Problem 9)
* Historically, at least part of the reason that we have adopted the base ten for our
number system is that humans have ten fingers; the implication of Problem 5.2, then,
is that the number of fingers thai Martians have is the base of their number system.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
147
Problem 5.3
A college student sent a postcard to her parents with the message
SEND
+ MORE
MONEY
If each letter represents a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), with
different letters representing different digits and the same letter
representing the same digit each time it occurs, how much money is being
requested? (SEND for the school tuition and MORE for other
expenses.)
Problem 5.4
Reconstruct the division
CECI
QUl)TROUVE
_E
0
{Qui trouve ceci? is French for Who finds this?) (M. Pigeolet, [60],
Vol. 2, p. 24)
POSITIONAL NOTATION
Our number system requires ten symbols—0, 1, 2, 3, 4, 5, 6, 7, 8, and
9—which we call digits. The position of each digit relative to the
decimal point indicates the "place value" of the digit—that is, by what
power of 10 it is multiplied. For example,
3 = 3
30 = 3
300 = 3
.3 = 3
1 = 3 • 10«
10 = 3 • 10^
100 = 3 • 10^
A=3- 10-1
and so on. (Note that 10^ = 1. In fact, 6^ is defined to be 1 for any
nonzero b.)

148
CHAPTER 5
Thus the number 24 represents 2 • 10 -f 4 = 2 • 10' -f 4 • 10^ while
the number 204 represents 2 • 100 -I- 0 • 10 + 4 • 1 = 2 • 10^ +
0 • 10' + 4 • 10«.
Note the importance of the digit 0 to help indicate position. This
valuable digit was first introduced by the Hindus, prior to the year 800.
In general, it is possible to use any positive integer b greater than
1 as a base. We need b symbols, 0, 1, 2, ..., 6 - 1, and we still make
use of positional notation.
Thus, in base two, we need two symbols, 0 and 1, and the number
1101 represents 1 • 2^ + 1 • 2^ + 0 • 2' + 1 • 2«. Similarly, in base three,
we need three symbols, 0, 1, and 2, and the number 120 represents
1 • 3^ + 2 -3' -f 0 • 3«.
In the base twelve, we need twelve symbols, 0, 1,2, 3, 4, 5, 6, 7,
8, 9, T, and E.
When more than one base may be involved, as in Problem 5.2,
it is necessary to indicate the base with respect to which a number
is being expressed. This is done by appending a subscript to the
number.
For example, 110 may be ambiguous but the following are not:
(llO)two = 1 -2^ + 1 -2' -f 0 •2«
(110)three= 1 ' 3^ + 1 ' 3' + 0 • 3«
(llO)ten = 1 • 10^ -f 1 • 10' -f 0 • 10«.
If it is clear from a discussion what base is intended, the
subscript may be omitted.
^i^ipp^p
CHANGING BASES
Although every integer may be expressed with respect to any base, the
representation of a number with respect to one base will usually differ
from its representation with respect to another base. For example.
and
(201)three=2 • 3^ -f 0 • 3 -f 1 = (19)ten
(lOOll)two = 1 • 2^ -f 0 • 2^ -f 0 • 2^ -f 1
= (19)ten
(34)five=3-5H-4 = (19)ten.
2 + 1
Thus, (201 )three, (10011 )two > (34)five > and (19)ten are different
representations of the same number—the number that we refer to as 19. In the
first representation, the 19 objects have been grouped in terms of
powers of three; in the second representation, powers of two have been
used; and in the third, powers of five.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
149
Since we are used to working in base ten, it is sometimes useful to
be able to convert from the representation of a number in base b to its
base ten representation, or vice versa.
To convert a number from base b to base ten is easy; it is simply
a matter of expanding the number in terms of powers of b and then
multiplying and adding. For example.
(1101101)two= 1 -2^
= 64
-^(109),,
-f 1 • 2^^ + 0 • 2^ + 1 • 2^ + 1
+ 32 +8 +4
2^+0-2^ + 1
+ 1
2«
1. [a] Convert each of the following to base ten.
(a)(1101)two (b)(1001)
seven (c)(3.5)s,x (d)(T81)
eleven •
2. Convert each of the following to base ten.
(a)(10011101),wo (b)(5403)six (C) (ET09),welve (d) (31.2)seve„.
PRACTICE
PROBLEMS
5.A
To convert from base ten to base b is slightly more complicated
because it requires division. We must express the given number as a
sum of multiples of powers of b.
We illustrate by giving one general technique on the left and a
corresponding example on the right:
If n is the given number that we
wish to write in base b, we begin
by finding the largest power of b
that does not exceed n.
Say this power is 6^<.
We next divide n by b^>, that
is, we use the division algorithm
(Theorem 4.7) to find q^ and r^
such that
n = q,b^^ -f r,,
with 0 < rj < 6^.. (Since
w < 6^' + •, it is easily seen that
q.<b- 1.)
Suppose we want to write
185 [= (185)ten] in base 3.
The powers of 3 are: 3^ = 1,
3^ = 3, 3^ = 9, V = 27, 3^ - 81,
3^^ = 243, ....
The highest required power is
3^ = 81, since 243 is too big.
That is.
so q.
-- 81
< 185 < 243 =
2
81)185
162
2, and r^ = 23
185
= 2-81+23.
V.

150
CHAPTER 5
We now repeat the process
using ri in place of n. That is,
finding the highest power of b
that does not exceed r^ (say, b^^)
we divide r^ by this power,
obtaining
r, = q^ b^^ + r^
where 0 < r2 < b^^.
Substituting in the expression
for w, we get
n = q^b^^ + ^2 ^^' + ^2 •
(Note that r^ < b""^ <r^ < b""^ and,
as above, 0 < q^ <b - \.)
Continuing in this manner, we
eventually have n expressed as a
sum of multiples of powers of b:
n = q^b^^ -\- q^b^^ ■¥ • • • + qkb''^
The process must terminate
with a remainder of zero, since
each remainder is smaller than
the previous one.
Remembering to include the
missing powers of b with
coefficient zero, we can now
express n in base b.
We continue, using 23:
Since V = 27 is too big, we
divide by V = 9.
2
9)23
18
~5
23 = 2 • 9 -f 5
185 = 2 -81 + 2 -9 + 5.
Continuing, using 5, we get
5=1-3 + 2
185 = 2 -81 + 2 -9 -f 1 • 3
+ 2-1
= 2 • 3^ + 2 • 3^ H- 1 • 3^
+ 2 • 3«.
Since the 3^ term is missing, we
add it in with coefficient 0:
185 = 2 • 3^ + 0 • 33 + 2 • 32
+ 1 • 3^ + 2 • 3«
= (20212)three.
PRACTICE
PROBLEMS
5.B
CZIJ
1. [a] Convert (2087)ten into each of the following bases:
(a) 2 (b) 3 (c) 6 (d) 7 (e) 12.
2. Convert (17854)ten into each of the following bases:
(a) 2 (b) 3 (c) 5 (d) 12 (e) 16.
3. [a] (a) Express two in base two.
(b) Express three in base three.
(c) If 6 is a positive integer greater than 1, express b in base 6.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
151
Aside from the historical significance, other bases are used in various
ways today. For example, many retailers count inventory in terms of
dozen (12) and gross (12^); they are essentially using base twelve.
Thus, for example, 4 gross 3 dozen and 11 items could be expressed
as (43E)tweive • For another example, we may consider time as being
measured in base sixty (a remnant from the Babylonian system). Thus,
is really
3 hours 12 minutes and 23 seconds
3 • 60^ + 12 • 60 + 23 seconds.
Probably the most important system today other than the decimal
system is the binary system—base two. Since an electric switch may
be placed in one of two states [off (0) or on (1)], it is possible to
represent a number in the binary system by turning a sequence of
switches on or off. Thus, operations with numbers may be carried out
electronically. This is the way in which some computers do
calculations. The binary system also has many applications to mathematical
puzzles. Problem 5.1 is one example.
If you have not already solved it, try Problem 5.1 (page 146)
now before reading on.
Problem 5.1 is attributed to Bachet de Meziriac (1581-1638). The
problem essentially says: What is the smallest number of weights that
can serve to counterbalance each integral number of grams from 1 to
some fixed number «?
The answer lies in the binary system.
Recall that every positive integer n may be expressed in the binary
system by first expressing w as a sum of multiples of powers of 2. Since
the base is 2, each coefficient is either 0 or 1. That is,
w = ^0 • 1 + a, • 2 + ^2 • 2^ -f a, • 2' -f • • • -f a/j • 2^
where each a, is either 0 or 1.
Suppose we had weights in the denominations of 1 gram, 2 grams,
2^ grams, and so on. Then, to balance an object weighing n grams, we
need only select those weights corresponding to powers of 2 that have
coeliftcient 1 in the expansion of n.
For example, the binary expansion of 109 is
109 = 1 • 2^ -f 1 • 2*^ -f 0 • 2-* + 1 • 2' + 1 • 2^ + 0 • 2^ -f 1 • 2".
Thus, an object weighing 109 grams can be balanced by placing 64,
32, 8, 4, and 1 gram weights in the other pan.
Solution of
Problem 5.1

152 CHAPTER 5
The fact that the coefficients in the binary expansion of a number
are at most 1 means that we need only one 2' gram weight for each i.
In the version of the problem at hand, the assayer wanted to be
able to balance all weights up to 127 grams. He therefore may select
1, 2, 4, 8, 16, 32, and 64 gram weights—seven weights in all. (In
general, to be able to balance all weights up to w, we may choose
weights of 1, 2, 4, ..., 2^ where 2^ < n < 2^^\)
The discussion above shows that the weights 1, 2, 4, ... will suffice,
but the question remains as to whether it might be possible to use
still fewer weights.
We demonstrate that this is not possible by showing that, if a set of
k weights suffices to balance every integral weight up to w, then n must
be less than 2^. In other words, if 2^ < n < 2^'^\ then we need at least
k -f 1 weights. (And since 1, 2, ..., 2^ will do, the minimum number
of weights needed is exactly k -\- \.)
Suppose now that we have k weights which can balance every
integral weight (in grams) up to n.
Since we can balance 1 gram, one of the weights must weigh 1 gram.
Call this weight W^.
Since we can balance 2 grams, we must have a second weight, W2,
which does not exceed 2 grams, that is, R^2 = 1 or 2. (If every weight
other than W^ weighed more than 2 grams, we could not balance an
object weighing exactly 2 grams.)
W^ and W2 together weigh at most 1 + 2 == 3 grams. Hence, to be
able to balance an object weighing 4 grams, we need a third weight,
R^3, weighing no more than 4 grams.
Since IFi, IF2, and W^ together weigh at most 1+2 + 4 = 7 grams,
we need a fourth weight, W^, not exceeding 2"* = 8 grams, in order to
be able to balance an object weighing 8 grams.
And so on. For each r, we must have another weight, Wr, that does
not exceed 2''~^ grams.
Thus, the sum of all the weights IFj, W2, ..., Wk cannot exceed
1 + 2 + 2^ + • • • + 2*- 1 = 2^ - 1 grams.*
Thus, n cannot exceed 2^ — 1; that is, n is less than 2^.
The argument above gives the outline of a proof, but the lack of
precision in the "and so on" step leaves something to be desired. The
proof can, with very little change, be formalized through the use of
mathematical induction. (A discussion of mathematical induction
appears in Appendix B.)
* If 1 + 2 + •• •+ 2* ' = X, then 2x = 2 + 4 + •• • + 2*, so x = 2x - x = (2 + 4 + • ••
+ 2*) - (1 + 2 + • • • + 2*"') = 2* - 1 (all other terms cancel). Or see Appendix B for a
proof by mathematical induction.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
153
Before leaving this version of Bachet's weight problem, one more
comment is in order. Suppose we wish to be able to balance all weights
up to 100 grams. From our discussion above, we know that at least
seven weights will be needed; and, in fact, the seven weights 1, 2, 4,
8, 16, 32, and 64 will do. However, since we wish to go up only to 100,
there are other sets of seven weights that will also work, for example,
1, 2, 4, 8, 16, 32, and 63; but no set of fewer than seven weights
will do. The first set will actually enable us to go up to 127, while the
second set will enable us to go up only to 126, but both sets will handle
weights up to 100.
Another version of the weight problem is also of interest—weights
may be placed on both pans of the balance. We leave this for an
exercise at the end of the chapter (see Exercise 5.25.)
ADDITION AND MULTIPLICATION
IN OTHER BASES
Just as we can add and multiply numbers in base ten, we can do so in
any base. Consider the following addition in base ten:
28
H- 57
+ 39
To obtain the sum, we say 8 plus 7 is 15, plus 9 is 24; so we write 4
under the 8-7-9 column and carry 2. Then we say 2 H- 2 is 4, plus 5
is 9, plus 3 is 12. We write 12 and thus obtain 124 as our answer.
The justification for this process is as follows:
28 + 57 -f 39 = (2 • 10 -f 8) -f (5 • 10 -f 7) + (3 • 10 + 9)
= 2- 10 + 5- 10 + 3- 10 + 8 + 7 + 9
= (2 + 5 + 3) • 10 + (8 + 7 + 9)
= (2 + 5 + 3) • 10 + 24
= (2 + 5 + 3) • 10 + 2 • 10 + 4
= (2 + 5 + 3 + 2) • 10 + 4 • 1 (notice 2 being carried)
= 12 • 10 + 4 • 1
= (10 + 2) • 10 + 4 • 1
= 1 • 10^ + 2 • 10 + 4 • 1
= (124)ten.

154
CHAPTER 5
Base ten arithmetic should be second nature to us, because the
addition and multiplication tables were pounded into our heads in the
early grades at school. Base b arithmetic would be just as easy if we
knew the base b tables. These tables can easily be formed by using
base ten equivalents. For example, in base five, we first write the base
five equivalents of the base ten numbers from 0 to 16*:
base ten
0 12 3 4 5 6 7 8 9 10 11 12 13 14 15 16
base five 0 1 2 3 4 10 11 12 13 14 20 21 22 23 24 30 31
This table can be used to add or multiply any two base five numbers
from 0 to 4, by doing the addition or multiplication in base ten and
then converting back to base five. For example.
3five H- 3fi
+ 3ten = 6t«
or
(3 • 4)five = 3ten ' 4ten = (12)t(
(ll)five
(22)five .
We can now construct the addition and multiplication tables for base
five (although it is no longer really necessary to do so). See Figure 5.2.
+
0
1
2
3
4
0
0
1
2
3
4
1
1
2
3
4
10
2
2
3
4
10
11
3
3
4
10
11
12
4
4
10
11
12
13
•
0
1
2
3
4
0
0
0
0
0
0
1
0
1
2
3
4
2
0
2
4
11
13
3
0
3
11
14
22
4
0
4
13
22
31
FIGURE 5.2
To add and multiply larger numbers in base five, we make use of
positional notation in following the same procedure as in base ten,
carrying and borrowing where necessary.
For example, in base five,
since
(4 + 3)five
(1 + 3 -f 2)five
34
H-23
112
= (12)five
= (ll)five
(write 2, carry 1)
(write 1, carry 1);
* We stop at 16 because, in base ten, 16 = 4 4. In general, we must go up to
(^- l)-(^- 1).

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
155
and
since
34
X 3
212
(3 • 4)five = (22)five
(3 • 3)five H- 2five = (21)five.
(write 2, carry 2)
The justification of this procedure is similar to that for base ten.
For example, in base five,
34 -f 23 = (3 • 5 + 4) -f (2 • 5 -f 3)
= (3 -f 2) • 5 -f (4 -f 3)
= 5-5H-(12)five
= 1 • 5^ -f 1 • 5 + 2
= (112)five.
1. [a] Construct the addition and multiplication tables for the base
eight.
2. Construct the addition and multiplication tables for each of the
following bases: (a) two (b) three (c) seven.
3. [a] Add : (a) (20 12)three + (2 122)three . (b) (222)three + ( 120)three •
4. Add: (a) (5621) seven + (3106) seven • (b) (4613)
seven H- (2536) seven •
5. [a] Subtract: (a) (101101101)two - (111111 l)two. (b) (lllllOO)two
-(lOllOOl)two.
6. Subtract: (a) (2713)eight - (1746)eight • (b) (4531)eight - (2617)eight •
7. [a] Multiply : (a) (20 12)three ' (20 1 )three . (b) (20 1 )three ' (20 1 )three •
8. Multiply: (a) (5621) seven
(3106) seven • (b) (2463)
seven
(253) seven •
9. Use binary system addition to show that 1 -f 2 H- 2' -f • • H- 2*
= 2^+1 - 1.
PRACTICE
PROBLEMS
5.C
We are now ready to solve Problem 5.2 (page 146). If you
haven't already solved it, try again now.

156
CHAPTER 5
Solution
of Problem
5.2
Consider the coefficients of the equation as being in some base, b.
The equation is
(5)bX^ -(50)bX-f(125)b = 0.
Converting to base ten, this becomes
5x^ - (5b -f 0)jc + (b^ + 2b + 5) = 0.
We are told that jc = 5 is a solution. Thus,
5 • 5^ - (5b H- 0) • 5 H- (b^ H- 2b H- 5) = 0.
125 - 25b + b^ + 2b + 5 - 0.
b^ - 23b + 130 = 0.
(b - 13)(b - 10) = 0.
b = 13 or b = 10.
Therefore, since Martians have more fingers than Earthlings have,
they must have 13 fingers. (We should check that in base thirteen, the
equation 5x^ - 50x H- 125 = 0 has two solutions:
(5)th irteen-X^ — (50)thirteen-X^ + (125)thirteen — 0.
Converting to base ten, we get
5x^ - 65jc + 200 = 0
x^ - 13jc + 40 = 0
(x - 5)(x - 8) = 0
X = 5, X = 8 in base ten, which also gives x = 5, x = 8 in base thirteen.)
CRYPT ARITHMETIC
The remaining problems discussed in this chapter fall into the general
area of recreational mathematics known as cryptarithmetic. This type
of problem was popularized during the 1930s in Sphinx, a Belgian
journal of recreational mathematics. Each problem is an arithmetic
calculation in which digits are either coded or partially missing. Our
task is to reconstruct the original calculation.
In many problems of this type, positional considerations are often
relevant. For example, if
AB
+ CD
EFG
then E must be 1. To see why, remember that the first written

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
157
digit of a number is never zero, so E is at least 1. Since neither AB
nor CD can be larger than 99, their sum is no larger than
99 -f 99 = 198. This shows that E is no larger than 1, so E = 1.
Similarly, if
ABCD
- EFG
HI
then A must be 1, B must be 0, and E must be 9 (otherwise the
difference between ABCD and EFG would be at least 100 and we would
have a three digit difference rather than just HI.)
We are now ready to attack Problem 5.3 (page 147). If you
have not already solved it, try it again.
Cryptarithmetic problems in which letters form meaningful words are
also called Alphametics, a term coined by J. A. Hunter [31]. Problem
5.3, one of the most famous cryptarithmetic problems, is such a
problem. All the information of the problem is given in code. (Note
the similarity with cryptograms, except that here each distinct letter
stands for a specific digit rather than for another letter.) The tools
needed to solve the problem are just the basic properties of base ten
arithmetic and a little ingenuity.
To describe our reasoning, we label the columns starting from the
right. That is, the D-E-Y column is the first column, and so on.
Observe that the M in MONEY must result from a "carryover," since
there are no other numbers in the fifth column. Hence, as we observed
at the beginning of this section, M must be 1 (SEND < 9999 and
MORE < 9999 so MONEY < 19998.) Substituting 1 for M
everywhere M occurs, we obtain
SEND
lORE
lONEY
Since END < 999 and ORE < 999, at most 1 can be carried over from
the addition in the third column.
Consider the fourth column. There are two possible cases:
Case 1
If nothing was carried from the
third column, then S -f 1 = IG
+ O. Since S < 9, this gives
S = 9 and O = 0.
Case 2
If 1 was carried from the third
column, then S -f 2 = 10 -f O,
or, equivalently, S = 8 + O.
Thus, S = 8 and O = 0, or S = 9
and O = 1. But M = 1, so O
cannot be 1. Hence S = 8, O = 0.
Solution
of Problem
5.3

158 CHAPTER 5
So O = 0 in either case, and S = 9 if there is no carryover from the
third column, and S = 8 if there is a carryover.
Substituting,
SEND
lORE
lONEY
Look now at the third column. Since at most 1 can be carried from
the second column (99 -f 99 < 198),
E-fO(+l) = N(-f 10).
The (H- 1) indicates the possibility that 1 was carried from the second
column, and the (-f 10) indicates the possibility that the sum of the
third column might be 10 or more. But the only way that E + 0 (+ 1)
could be N + 10 is if E = 9 and N = 0, which cannot be (O = 0).
Hence, there is nothing carried from the third column. Thus, 8 = 9
and E (+ 1) = N.
Since E # N, we must have a carryover from the second column.
Thus,
E + 1 = N
and
N + R (+ 1) = 10 + E.
Substituting E + 1 for N in this last equation,
E + 1 + R (+ 1) = 10 + E
R(+ 1) = 9.
Since S = 9, we must have R = 8 and there must be a carryover from
the first column. Thus
9 END
1 0 8 E
lONEY.
with
E + 1 = N
and
D + E = 10 + Y.
Since D is at most 7 (R = 8, S = 9) and Y is at least 2 (0 = 0,
M = 1), E must be at least 5, since D + E = 10 + Y.
Also, since E + 1 = N < 7, then E < 6.
If E = 6, then N = 7, and so D can be at most 5; but then
D + E < 11, which cannot be.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
159
Hence, E = 5, N = 6, D = 7, and Y = 2. That is,
9567
+ 1085
10652.
The student is asking for $10,652!
1. [a] Consider the following cryptarithmetic problem (distinct letters
represent distinct values):
CF
AB)CDEF
AB
BGF
BGF
PRACTICE
PROBLEMS
5.D
(a) What are the possible values for C?
(b) Which is larger, B or F? (Note F AB = BGF.)
(c) What are the possible values for B and F? (Note F • AB ends
in F.)
(d) For each possible set of values for B and F, what are the possible
values for A? (Note F AB = BGF.)
(e) Which of the above cases actually lead to a solution? What is the
solution ?
Consider the following cryptarithmetic problem (distinct letters
represent distinct values):
ABC
X BC
EDC
FEB
CDAC
(a) What are the possible values for C? (Note, C • C ends in C.)
(b) For each possible value for C, what are the possibilities for B?
(B C ends in B.)
(c) Eliminating those values of C for which no values of B exist,
what is the unique remaining possibility for C?
(d) What can F be equal to? (Note addition in the fourth colurrm.)
(e) For which of the remaining possible values for B is it possible to
determine a value for A? (ABC B = FEB.)
(f) Complete the solution.

160
CHAPTER 5
Solution of
Problem 5.4
It is now time to try Problem 5.4 (page 147) again, if you
have not already solved it.
Problem 5.4 is a partial ghost—we are told which positions are
occupied by numbers, but we are given only partial information about
what each letter stands for. (As before, different letters represent
different numbers, and each time a letter appears it represents the same
number.)
To facilitate referring to specific entries, we label the rows of the
division as follows:
quotient
divisor) dividend
row 1
row 2
row 3
row 4
row 5
row 6
row 7
We begin the solution by observing that the last entries in rows 2,
4, and 6 respectively must be U, V, and E, since these are brought
down from the dividend.
CECI
QUI)TROUVE
U
__V
__E
__E
We next observe that the final entry of row 7 must be E, since
there is no remainder. Since the final entry in row 5 is also E, we
conclude that C • I ends in E and P also ends in E. Making a chart
of the possible values of I and recalling that C, E, and I are distinct,
we are left with only three possible cases:

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
161
/
0
1
2
3
4
5
6
7
8
9
E
impossible
impossible
4
9
6
impossible
impossible
9
4
1
C
7
impossible
9
impossible
3
impossible
I
2
2
8
E
4
4
4
C
7
7
3
Q
1
1
1
u
0
3
0
The second of the possible cases may be eliminated because, if
C = 9, then row 1 = C • QUI > 900, which forces T to be 9; but T
and C cannot both be 9.
In the remaining two cases, since C • QUI = row 1 and I • QUI =
row 7 are three-digit numbers, Q must be 1 and we have the
possibilities for U shown in Figure 5.3.
(U ^ 1, 2, or 4, and 152 • 7 is four digits.;
(U ^ 1, and 8 • 128 is four digits.)
FIGURE 5.3
Since there is no remainder, row 6 must equal row 7 = 1- QUI.
But row 6 is obtained by subtracting row 5 from row 4; hence, the
second entry in row 6 is V (-f 10) - E (here the (-f 10) indicates the
possibility of borrowing). This enables us to compute V. We do so in
each case (Figure 5.4):
impossible V = E = 4
impossible V = U = O
I
2
2
8
E
4
4
4
C
7
7
3
Q
1
1
1
U
0
3
0
row 6
204
264
864
V
4
0
0
FIGURE 5.4
In the first case, V = E = 4; and, in the third case, V = U = 0.
Hence, these cases may be eliminated.
We are left with Q = 1, U = 3, I = 2, C = 7, E = 4, and V = 0.

162
CHAPTER 5
It is now a simple matter to complete the problem. We first find rows
1, 3, 5, and 7 (and hence 6) by multiplying C • QUI, E • QUI,
C • QUI, and I • QUI respectively. This gives
7472
132)TRO304
9 2 4
_ _3
5 28
__0
924
264
264
Then, working from the bottom up, we can fill in row 4, then row 2,
and finally the dividend. We obtain
7472
132)986304
924
~623
528
~950
924
~264
264
Thus, T = 9, R = 8, 0 = 6 and, as we already found, Q = 1, U
I = 2, C = 7, E = 4, V = 0.
3,
Problems 5.3 and 5.4 involve base ten arithmetic, as do most of the
cryptarithmetic exercises at the end of this chapter. It is also possible,
however, to consider cryptarithmetic problems in other bases—see, for
example. Exercises 5.57-5.64.
THE CHAPTER IN RETROSPECT
This chapter first considered some aspects of base ten (decimal)
arithmetic, then generalized to arithmetic in an arbitrary base b.
(Mathematicians love to generalize.) We found that no matter what
base we work in, the operations are essentially the same.
The problems in the text have stressed some recreational aspects of
numerical computation. The exercises at the end of this chapter

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
163
continue along the same lines. Some of the games we will encounter
in Chapters 7 and 8 will also make use of the concepts we have
considered here.
Exercises
Some of the concepts discussed in Chapter 4
can be helpful in the solutions of some of the
problems in this chapter. These exercises are
denoted by #.
Decimal Notation
In Exercises 5.1 through 5.14, all numbers are
assumed to be expressed in the base ten.
5.1. [a] In his autobiography, Count Burr
Turr tells about an occasion during his
childhood when he decided to write all the counting
numbers from one to one million. It was a
noble undertaking, but his arm gave out after
he had written only 31,676 digits.
Assuming that the Count was using decimal
notation, what was the last digit he wrote
before his arm grew numb?
# 5.2. (a) [h] Show that the last digit of the
square of any integer must be 0, 1, 4, 5, 6,
or 9.
(b) [a] Which digits could possibly be the
last digit of the cube of a positive integer?
(c) Which digits could possibly be the last
digit of the fourth power of an integer?
# 5.3. [s] Show that the next to last digit of
a perfect square (a number that is the square
of a positive integer) is odd if and only if the
last digit is 6.
# 5.4. [h] [a] For the first time in the
history of Weapons, Armaments, and Rockets,
Inc., there was a real battle over the presidency
of the company. Each of the five candidates
had the backing of more than 100,000 shares of
stock. The final tabulation was recorded on a
piece of paper; but, before the results could be
read at the board meeting, someone spilled
some coffee. The resulting situation is shown
in Figure 5.5.
46 M. Addison
235 A. Jax, Jr.
168 L. Cohn
44 R. Velt
174, 329 W. Wills, Jr.
FIGURE 5.5
Fortunately, someone had noted earlier that
the number of votes received by the winner
was a perfect square.
Who won the election?
5.5. An automobile riding along Route 101
passed a sign showing the distance to Los
Angeles. This distance, in miles, was a three-
digit number, the middle digit of which was
zero. Exactly one hour later, the car passed
another sign showing the distance to Los
Angeles. This time, the distance was a two-
digit number, containing the same two nonzero

164
CHAPTER 5
Mm,
digits as the original sign, but in the reverse
order. Exactly one hour after that a sign
indicated the distance to Los Angeles as a
two-digit number containing the original two
nonzero digits in their original order.
Assuming that the car maintained a uniform
speed, how long after passing the third sign
did the car arrive in Los Angeles? ([60],
Vol. 5, p. 139)
# 5.6. [s] [a] Find all two-digit numbers
that are divisible by the product of their
digits.
# 5.7. Find all two-digit numbers that are
divisible by the sum of their digits.
# 5.8. [h] Find all two-digit numbers that
have two distinct nonzero digits, such that the
difference between the number and its reversal
(the same number with the order of the digits
reversed) is divisible by the sum of the digits of
the number.
# 5.9. (a) Find all two-digit numbers that,
when added to their reversals, give perfect
squares. ([45], Problem 125e)
(b) [s\ [a] Find all two-digit numbers that,
when subtracted from their reversals, give
perfect squares.
(c) [a] Find all two-digit numbers that,
when subtracted from their reversals, give
perfect cubes.
# 5.10. Find all two-digit numbers with
nonzero digits, that, when added to their
reversals, give a result divisible by 66.
# 5.11. [h] [a] Find two-digit numbers with
distinct digits such that the difference
between the square of the number and the
square of the reversal of the number is itself
a perfect square. ([60], Vol. 1, p. 85)
# M 5.12. Show that if any three-digit
number is subtracted from its reversal, then
the result is divisible by 99.
5.13. [s] [a] The magician removed a sealed
envelope from his pocket and handed it to a
member of the audience. He then asked
another member of the audience to select a
three-digit number with distinct digits, to
reverse the order of the digits, and to subtract
the smaller three-digit number from the larger.
The magician then asked his subject whether
the resulting number had three digits. When
he was told that it did, he asked the subject
to reverse the order of the digits in the result
and to add the number thus obtained to the
previous result. The subject was then told to
announce the sum to the audience. After he
did so, the sealed envelope was opened; inside
it was an index card on which the subject's
result was correctly predicted.
How did the magician do the trick?
5.14. [a] The magician asked the subject to
do the following:
1. Represent the month of your birth as
a number. For example, January = 1,
February = 2, and so on.
2. Multiply that number by 5.
3. Add 17.
4. Double the result.
5. Subtract 13.
6. Multiply the result by 5.
7. Subtract 8.

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
165
8. Double the result.
9. Add 9.
10. Add the day of the month on which
you were born. For example, if your
birthday is August 23, then, in step 1, start
with 8 and, in step 10, add 23.
11. Tell me the result.
The subject announced: 1332.
"Oh, I see your birthday was this past
Thursday," said the magician.
If today is Tuesday, what is today's date?
Binary System
5.15. [s] [a] Tweedledee and Tweedledum
are playing a game. Tweedledee thinks of a
number between 1 and 1000, and Tweedledum
tries to guess the number. Tweedledum is
allowed to ask a limited number of questions
of the form " Is the number greater than (less
than) so and so?"
How many questions must he be allowed to
ask in order for him to be sure of getting the
right number on his first guess if he proceeds
properly? How should he proceed?
5.16. [h] [a] Now it is Tweedledum's turn to
think of a number; but the rules have
changed slightly. Tweedledee is allowed eight
guesses. With each incorrect guess.
Tweedledum says either "higher" or "lower,"
whichever is correct.
What is the largest number that
Tweedledum should be allowed to pick if Tweedledee
is to be assured of success provided that he
proceeds properly? How should Tweedledee
proceed ?
5.17. Once again the rules have changed (see
Exercises 5.15 and 5.16). This time,
Tweedledee may choose any positive integer
whatsoever, and Tweedledum is allowed as many
"yes or no" questions as he likes; but, if he
receives two "no" answers, then Tweedledum
loses.
How should Tweedledum proceed to be
sure of success ?
5.18. [h] [a] The magician selected three
volunteers from the audience. He instructed
each to think of a number between 1 and
31. He then displayed five cards containing
numbers as indicated in Figure 5.6. He asked
each volunteer to determine which cards
contain the number he or she selected.
Card A
1
3
5
7
9
11
13
17 25
19 27
21 29
15 23 31
Card B 1
2
3
6
7
10
11
14
18 26
19 27
22 30
15 23 31
Card C
4
5
6
7
12 20 28
13 21 29
14 22 30
15 23 31
Card D 1
8 12 24 28 1
9 13 25 29
10 14 26 30
11 15 27 31
1 Card E
1 16 20 24 28
17 21 25 29
18 22 26 30
19 23 27 31
FIGURE 5.6
"A, C, and E," announced the first
volunteer.
"Then your number is 21," replied the
magician quickly.
"My number is only on card D," said the
second volunteer.
"Then your number is 8."
"My number is on cards B, C, D, and E,"
proclaimed the third volunteer.

166
CHAPTER 5
"Then your number is 30," the magician
announced triumphantly.
How was the magician able to determine
the selected number so quickly? And why
does the trick work?
5.19. The magician displayed a packet of 16
cards, each bearing a number from 0 to 15.
The pack was arranged, from top to bottom,
in numerical order. He then dealt two piles
face up on the table as follows: First he dealt
the 0 face up and then 1 face up in another
pile. Then he dealt 2 face up on top of 0, 3
face up on top of 1, 4 face up on top of 2,
and so on. When he finished dealing the 16
cards, he selected the pile into which the last
card was dealt, placed it face down in his left
hand, and then placed the remaining pile face
down on top of the one in his hand. He thus
again held the 16 cards face down in his hand,
although they were no longer in their original
order. He then repeated the entire process,
alternately dealing to two piles of face up
cards and then picking up the two piles in the
prescribed manner. After he had completed
this process a total of four times, he displayed
the cards and they were back in their original
order.
(a) [s] Why does this work?
(b) [a] What would happen if he went
through the above process four times, dealing
the cards face down rather than face up?
Explain why.
(c) [a] What would happen if he dealt the
cards face down, but placed the pile into
which the last card is dealt on top of the
other pile rather than below it?
(d) For each of the three methods of dealing
described above, how many times would the
process have to be repeated for the cards to
return to their original order, if the packet
contains 32 cards rather than 16?
Other Bases
'It 5.20. [h] The magician placed an apple, a
banana, a peach, and 25 matchsticks on the
table. He then selected three volunteers from
the audience. Their names were Alvin, Julia,
and Melonie. The magician then handed one
of the matchsticks to Alvin and two match-
sticks to Julia, but he did not give any to
Melonie. He was then blindfolded, and each
of the volunteers was asked to remove one of
the fruits from the table.
''Whichever of you removed the apple
should now remove a number of matchsticks
equal to the number I handed you a moment
ago. Whoever removed the banana should
remove three times as many matches as you
were originally given. Finally, the remaining
person should remove nine times as many
sticks as I gave to him or her."
"Now, tell me how many sticks remain on
the table."
"Seven," was the reply.
"Then Alvin took the peach, Julia took the
banana, and Melonie took the apple," declared
the magician.
Naturally, he was correct.
Explain how this trick works.
5.21. The magician held up a packet of 27
cards, marked, in numerical order from top to
bottom, with the numerals from 0 to 26.
He then dealt them face up into three piles
as follows: First he put 0, 1, and 2 in three
separate piles face up on the table. Then he
put 3 on 0, 4 on 1, 5 on 2, 6 on 3, 7 on 4,
and so on.
When he finished dealing, he picked up the

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
167
pile into which the last card was dealt and
placed it face down in his left hand. Then he
placed the next pile face down on top of the
pile in his hand; and, finally, he placed the
third pile face down on top of the other two.
He then repeated the entire process two
more times. When he was done, the cards
were back in their original order.
Explain why this happens. (See also
Exercise 5.19.)
5.22. The magician removed a packet of 27
cards from an ordinary deck of playing cards.
He handed the packet to a volunteer from the
audience and asked her to select a card, look
at it, and then mix it up in the packet. The
magician then asked the audience for a
number between 1 and 27.
"Eight," someone called out.
Next the magician dealt the 27 card packet
face up into three piles of nine cards each and
asked the volunteer to indicate which pile
contained the selected card. The subject
pointed, and the magician picked up the cards,
placing them face down in his hand with the
indicated pile between the other two. He then
again dealt three face up piles, alternating, one
card at a time, between the piles (as in Exercise
5.21). The subject again indicated which pile
contained the selected card, and the magician
picked up the cards, this time placing the
selected pile face down below the other two
piles.
Once more the magician dealt three piles as
before, and this time, when he picked the
cards up, he placed the indicated pile face
down on top of the other two.
"Now," he said, "the audience has selected
the number eight."
He counted seven cards off the top of the
packet, and turned over the eighth card,
saying, "This is the card you selected." As
usual, he was correct.
(a) What would the magician have done
differently if the audience had selected the
number twenty-three?
(b) [s\ Explain why the trick works.
5.23. At a party celebrating the end of the
math course, a student who had decided to
major in magic asked one of her friends to
think of a number between 1 and 26.
She then placed the following cards on the
table and asked the friend which cards
contained the selected number.
Card A
1 10 19
4 13 22
7 16 25
Card B
2 11 20
5 14 23
8 17 26
Card C
3 12 21 '
4 13 22
5 14 23
CardD
6 15 24
7 16 25
8 17 26
Card E
9 12 15
10 13 16
11 14 17
Card F
18 21 24
19 22 25
20 23 26
"B, C, and E," was the reply.
"Then your number is 14."
(a) [a] What would the student have said if
her friend had announced "A, C, and F"?
(b) Explain why the trick works. ([60], Vol.
7, p. 96) (See also Exercise 5.18.)
M 5.24. \s\ Another student at the same
celebration produced three cards from his
pocket.
Card A
1
16
5*
17*
4
19
8*
20*
7
22
11*
23*
10
25
14*
26*
13
2*
Card B
2
13
6*
16*
3
20
7*
23*
4
21
14*
24*
11
22
15*
25*
12
5*

168
CHAPTER 5
Card C
5
10
15*
19*
6
11
16*
20*
7
12
17*
21*
8
13
18*
22*
9
14*
2. No more than four people are to receive
the same amount of money.
When Mr. Tiptoe died, his estate was valued
at $1,013,652.
How many of Mr. Tiptoe's heirs actually
received money, and how much did each
receive ?
He asked his subject to select a number
from 1 to 26 and to indicate those cards on
which the selected number appeared. In
addition, for each appearance of the selected
number, the subject was to say whether or
not the number was followed by an asterisk.
"Card A with an asterisk and Card C
without one."
"Your number is 8."
Explain how the trick works. ([60], Vol. 7,
p. 97)
¥ 5.25. [h] [a] In Problem 5.1 at the
beginning of the chapter, the assayer always placed
his ore sample on the left pan and weights
on the right pan. Normally, however, we may
place weights on both pans of the balance.
Thus, for example, with weights of 1 gram and
3 grams, we may weigh an ore sample weighing
2 grams—simply place the sample and the
1 gram weight in one pan and the 3 gram
weight in the other pan.
Assuming that the assayer uses both pans in
this manner, what is the smallest number of
weights he can possess and still be able to
weigh all possible ore samples weighing an
integral number of grams between 1 and 13?
Between 1 and 40? Between 1 and 121?
Generalize.
5.26. According to the will of G. Barefoot
Tiptoe, the value of his entire estate was to be
divided among his heirs (in a given order of
priority) according to the following rules:
1. Each heir is to receive $1, $5, or a
number of dollars equal to some other
power of 5.
5.27. [a] If 1/6 of 30 is 4, what is 1/4 of 10?
([41], Vol. 31, p. 178)
¥ 5.28. [h] In what base is 11111 a perfect
square? ([46], Vol. 15, Problem 149)
-¥ 5.29. [h] [a] In what base is 297 a factor of
792? ([41], Vol. 38, Problem 124)
5.30. [a] An integer is said to be even if it is
divisible by two, and is said to be odd
otherwise.
(a) If ABCDE is the base b representation
of a number w, under what circumstances is
n odd if
(i) b = ten (ii) b = two (iii) b = twelve
(iv) b = three (v) b — five?
(b) Generalize. That is, given the base b
representation of an integer «, how can you
quickly tell whether or not n is odd? ([46],
Vol. 12, Problem 197)
5.31. (a) [s] Prove that (121)^ is a perfect
square for every base b > 2.
(b) Prove that (1331)^, is a perfect cube for
every base b > 3. ([35], p. 52)
♦ -¥ 5.32. (a) [s] Prove that for any base
b > 2, the base 6 representation of the numbers
2(b — 1) and (b — 1)2 are the reversals of each
other.
(b) For any base 6 > 3, if the digits of the
representation of 3(6 — 1) are reversed and
the digit (6 — 1) is added at the right, prove
that the resulting three-digit number is the
representation of (b - 1)^ ([60], Vol. 2,
Problem 79)

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
169
Cryptarithmetic
5.33. [a] If P = ME, but I^ = YOU, what is
the reversal of the difference between ME
and I?
5.34. [A] Find MAID if O^ = DAD and
(IMy = MOM.
5.35. [h] [a] Albert M. Adic was brilliant in
all matters concerning numbers, but he was
sometimes careless about his spelling. Once he
wrote "to" instead of "two" in answer to an
age problem. His sister, who was always
teasing her brother about his spelling and his
love for problem solving, delighted at the
opportunity to correct him.
" Can't you spell? It's age ' two' not' to.' All
you ever do is math problems. You're a real
square."
" Oh yeah," replied Albert after a moment's
thought. "If AGE, TWO, NOT, and TO are
perfect squares, then do you know what you
are? You are a TWO + TO + TOO."
What was Albert calling his sister?
5.36. An archeologist discovered the
following addition and multiplication problems
engraved on the wall of a Roman ruin.
LIX
+ LVI
CXV
X^
Being a fan of cryptarithmetic problems, the
archeologist decided to consider the engraving
in that light. She found there is a unique
solution. What is it? ([38], Chapter 7,
Problem 29)
5.37. [a] Solve:
AHBCF
+EBFAF
AGGHFE
5.38. Shortly after Scott Troupe became the
new scoutleader in Camptown, his wife gave
birth to a baby boy. The scouts threw a
party in the baby's honor, with music provided
by the scout brass band. The local newspaper's
coverage of the gala proceedings was
headlined as follows:
TROOP
TOOTS
NEWSON
When Scott noticed the article, he
realized that it makes an interesting
cryptarithmetic problem: "TROOP + TOOTS =
NEWSON."
Can you solve it?
5.39. [s] [a] Solve:
LET
X NO
SOT
NOT
FRET
([35], Problem 5, p. 82)
5.40.|[h] [a] Solve:
ABC
DEC
FBGC
FHCE
GDE
HBHIC
5.41. Solve:
ABC
DEC
FGH
lAC
ACAE
AFAGCH

170
CHAPTER 5
5.42. Solve:
ABC
X ABC
DEFC
DAGH
DRAG
DDIHFG
5.43. [h] Fill in the missing digits:
0_
7_
_23_
_4 5_
5.44. [h] If all the X's represent the same
digit, find X and fill in the other digits of the
following multiplication problem:
X
X
_xx
X
# M 5.45. [h] [a] Ifall the X's represent the
same digit, find X and fill in the other digits
of the following multiplication problem:
♦ 5.47. [h] [a] In 1978, the authors
composed the following problem:
CHE I N
19 7 8
4 4 _
_ R
AVE _ BACH
Solve it.
5.48. [h] Fill in the missing digits:
-)-
([35], p. 81)
5.49. Fill in the missing digits:
_7
— )
XXX
# M 5.46. [h] [a] Solve:
AB
X CD
FEE
FAG
BFEE
([60], Vol. 1, p. 52)

MORE ABOUT NUMBERS: BASES AND CRYPTARITHMETIC
171
5.50. [a] Solve:
GEABF
ABC)ACDEEDFB
ACFD
CED
ABC
5.51. Solve:
HHF
I HP
GGB
GOB
FDAG
ABC)DECFGAI
DFJC
BDG
FBI
KAGA
KKEC
KCEI
KCEI
M M 5.53. [h] [a] Fill in the missing digits
(all the 4's are shown):
-)-
4_
l4
_4_4
4
M M 5.54. Don Dakota loved cryptic
division problems, so he asked a friend to take any
long division problem and to replace each digit
by a dash. The result was
J.
M 5.52. [h] [a] Find the missing digits (all
the O's and Ts are shown):
1
.1)101010101
_1_
1 1
1
__0
TZIl
1
._0
.1_
~\
1
0
Don's efforts to recapture the original
problem are doomed to failure because there is not
enough information to determine a unique
problem from which the ghost could have
come. Let us aid Don a little:
DOOMED
don)d I V I s I on
_ _ 4
([60], Vol. 1, p. 108)

172
CHAPTER 5
Now the problem has a unique solution.
Find it.
-¥ 5.55. [a] Fill in all the digits, given that
the quotient in the first division is the
dividend in the second.
.)-
-)-
([11], Problem 150)
# MM 5.56. [h] [a] Fill in the missing
digits:
-)-
# ♦ 5.58. [s] [a] In base seven, find all
squares of the form ABCABC. ([60], Vol. 4,
p. 152)
# M 5.59. [h] Show that there are no
squares of the form ABCABC in base five, if
A, B, and C represent distinct digits. ([60],
Vol. 4, p. 152)
M 5.60. [a] In what base(s) does the
following addition have a unique solution?
TEN
+ NOT
NINE
MM 5.61. [h]|[a] In what base(s) does the
following addition have a unique solution?
ONE
+ TWO
NEXT
M 5.62. If, in base nine.
TO
- BE
OR
all solutions.
and
NOT
- TO
BE
M M 5.63. [a] In what base does the
following division have a solution? Find the solution.
([11], Problem 145)
-)-
.0.
Cryptarithmetic in Other Bases
M 5.57. \h\ Given that, in base six,
GALON = (GOOy
and
ALONG = (OOGY
find the numerical value of each letter. (A.
Colago, [60], Vol. 4, p. 136)
([63], Problem 173)
M 5.64. Prove that for all bases b > 3,
(ONE)6+, - (ONE)6 = (ON)26+i, no matter
what digits O, N, and E represent, O # 0.

Solve It With Networks,
An Introduction
to Graph Theory
There are areas of mathematics whose development was motivated by
the desire to solve problems originally posed as recreations. As the
theory in these areas was further developed, important applications to
other fields were found.
One such area of mathematics is graph theory, which today has wide
applications in the design and programming of computers, and in life
sciences, electrical networks, and business, as well as other fields.
Although graph theory did not come into its own until the middle
1800s, the groundwork of this theory was laid by Leonhard Euler
(pronounced "oiler") in 1735. He had visited the town of Konigsberg,
Prussia, where he encountered the following problem that has become
known as the Konigsberg Bridge Problem:
In Konigsberg, there are two islands surrounded by the two branches
of the Pregel River. There are seven bridges crossing the river in
various locations in the city (see Figure 6.1). Can a person plan a
walk in the city so that he or she will cross each bridge exactly once?
A
FIGURE 6.1
173

174
CHAPTER 6
We use this problem as well as the remaining sample problems
below to introduce some of the basics of graph theory. As usual, we
suggest that you try them before reading on.
SAMPLE
PROBLEMS
Problem 6.1
(a) Solve the Konigsberg Bridge Problem stated above.
(b) Suppose that, owing to severe rusting, bridge b collapsed, leaving
the situation pictured in Figure 6.2.
FIGURE 6.2
Can a person plan a walk so that he or she will start at D and cross
each of the bridges once but not more than once? What if the individual
starts at B? Can a walk be planned that crosses each bridge exactly
once and terminates at the same place at which it starts?
(c) Suppose that the town planners want to build a new bridge to
replace the one that collapsed. However, instead of building it in the
same location as before, they wish the bridge to be situated in such
a way that, after it is completed, it will be possible for a person,
starting anywhere in town, to plan a walk that will cross each bridge
exactly once and return to the starting point.
Where should the new bridge be built?
Problem 6.2
Figure 6.3 shows the floor plan of a house having two exterior doors.
Is it possible to enter the house through door a, travel through the
house, passing through each doorway exactly once, and then exit
through door i?
Problem 6.3
Starting at point A in Figure 6.4, travel along the lines of the figure
so that each labeled point is passed through exactly once and you end
your trip back at A.

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
175
c
1
g
L
1
d
1
b
e
f
h
1 1
FIGURE 6.3
FIGURE 6.4
Problem 6.4
Three missionaries were conducting three cannibals to the mission
school. On the way, they came to a piranha-filled river, which had to
be crossed. There was a canoe that could hold only two people, but
only one of the missionaries and one of the cannibals could be trusted
to paddle without tipping it. In addition, as the cannibals had not yet
been shown the light, the missionaries were afraid to create a situation
in which cannibals would outnumber missionaries on either shore.
How should they cross the river?
GRAPHS
Euler's approach to the Konigsberg Bridge Problem was to realize
that the sizes of the land masses and bridges are not significant;
and so there is no loss of generality if we shrink the land masses to
points and represent the bridges as lines or curves connecting these
points. Doing this with Figure 6.1, we get Figure 6.5a. Similarly,
Figure 6.2 becomes Figure 6.5b.
FIGURE 6.5

176
CHAPTER 6
The new diagrams are called graphs or networks. The word graph
is being used here in a sense that may be unfamiliar to you. As used
in this chapter, a graph ^ consists of a finite set Y of points, called
vertices (vertex is the singular) or nodes, and a finite set S of line
segments or curves, called edges, branches, or roads. Each edge
either connects two distinct vertices (^^ ) or else it connects a vertex
to itself (0 ). In the latter case, the edge is called a loop.*
Figure 6.6 gives some examples of graphs and networks—highway
maps, diagrams representing chemical compounds, and electrical
circuits. The tree diagrams discussed in Chapter 1 are further examples.
H
H —C —C^
I \
NH2
Glycine
(b)
FIGURE 6.6
OH
Electrical circuit and its graph
(c)
* Present practice favors using the term graph for structures with at most one edge
between any two vertices and not containing any loops. The term multigraph is used for
structures that contain multiple edges (CZ>) or loops. The term network is used for
graphs whose edges are labeled with numbers representing physical quantities such as
distances or resistances.
Nevertheless, the use of these terms is not uniform, and in the interests of simplicity
and clarity, we use the terms graph and network synonymously, as defined above.

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
177
Many graphs, such as those in Figures 6.5a and 6.5b, can be drawn
on a piece of paper—that is, in the plane—so that edges intersect
only at vertices. A graph that can be drawn in this manner is called
planar.
There are also nonplanar graphs. For example, consider a graph
having five vertices, each pair of which is connected by exactly one
edge. (This graph is called the complete graph on five vertices.)
This graph is nonplanar. If we try to draw this graph on paper—in
the plane—we may start as in Figure 6.7.
Now, there is no way to connect B to E without intersecting one of
the edges already drawn. We might try to start over again but, no
matter how we begin, we will not succeed in drawing the entire graph
in the plane without introducing points of intersection other than the
vertices. On the other hand, if we consider the graph in three
dimensions (3-space), we can connect B to E by an edge that passes
over edge CD and does not intersect it.
Similarly, a graph consisting of two sets of three vertices each,
with edges connecting each vertex in the first set with each vertex in
the second set, cannot be drawn in the plane without introducing new
points of intersection. (For example, in Figure 6.8, C cannot be
connected to X in the plane, but may be connected to X in 3-space.)
This graph is called the complete bipartite graph on two sets of
three vertices. In general, a bipartite graph is one in which the
vertices can be divided into two sets so that each edge of the graph
connects a vertex in one set to a vertex in the other.
Although not all graphs are planar, it can be shown that all graphs
can be constructed in 3-space so that edges intersect only at vertices.
On the other hand, because all pictures have to be drawn on a two-
dimensional page, we must somehow represent nonplanar graphs in the
plane. In Figure 6.9 we illustrate two ways in which this can be done
A B A B
(a)
FIGURE 6.9
for the complete graph on five vertices. This book uses the approach
of Figure 6.9b. Note that, in this figure, the only vertices are A, B, C,
D, and E. The other points at which lines intersect are not considered
to be vertices or intersection points; they are only "apparent points of
intersection." With this in mind, the complete bipartite graph on two
sets of three vertices may be drawn as in Figure 6.10.
FIGURE 6.7
FIGURE 6.8
FIGURE 6.10

178
CHAPTER 6
D c C
FIGURE 6.11
EULERIAN PATHS AND CIRCUITS
If we start at a vertex in a graph, follow an edge from that vertex to
a second vertex, then follow an edge from that vertex, and so on;
and if no edge is followed more than once in this process, then the
resulting sequence of vertices and edges is called a path.
We may denote a path by listing the vertices and edges in the order
in which they are encountered. For example, the path in Figure 6.11,
starting at A and ending at C, can be denoted by
AaBeDcC.
If there is only one edge connecting two vertices, then omitting
specific mention of the edge should create no confusion. That is, the
path above could also be denoted by
ABeDC.
A path that terminates at the same vertex at which it starts is
called a circuit. A path that traverses every edge of a graph exactly
once is called an Eulerian (pronounced "oil air' ee en") path; an
Eulerian circuit is defined analogously; it is a circuit in which every
edge of the graph is traversed exactly once. Thus, for example,
DEABCFEBFDC is an Eulerian path for Figure 6.12a, and ABCDA
is an Eulerian circuit for Figure 6.12b.
A
(a) (b)
FIGURE 6.12
A graph possessing an Eulerian path is said to be traceable: By
following the path, we can trace the graph without repeating any
edge and without lifting our pencil from the paper.
Problem 6.1 (page 174) may now be rephrased as follows:
(a) Is the graph in Figure 6.5a traceable?
(b) Is there an Eulerian path starting with D for the graph in
Figure 6.5b? What about starting at B? Does the graph in question
have an Eulerian circuit?
(c) Can one edge be added to the graph in Figure 6.5b so that the
resulting graph will have an Eulerian circuit?
We can answer these questions if we can answer the following, more

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
179
general one: Under what circumstances does a graph contain an
Eulerian path (circuit) starting at a particular vertex?
Before we attempt to answer that question, let us consider some
further examples.
The graph in Figure 6.13a is clearly not traceable, because no path
can contain both edge e and edge f. The problem with this graph is
that it is not connected—it consists of two disjoint components.
(Mathematically speaking, a graph is said to be connected if, for every
possible pair of vertices, U and W, there is a path from U to W.)
To avoid problems of this type, we restrict our attention to connected
graphs.
A B A B
(a)
FIGURE 6.13
The graph in Figure 6.13b is connected, yet it is still not traceable.
There are three roads leading to D but because A, B, and C are all
dead ends, at most two of the roads to D can be traveled—one to
reach D and one to leave.
The addition of an edge from A to B in Figure 6.13c allows us to
find Eulerian paths: CDABD and CDBAD. That is, reaching D from
C, we can travel to A or B and still be able to traverse the third road
to D. As the reversal of an Eulerian path is also an Eulerian path,
we also have DBADC and DABDC.
All four of these paths begin at C or D and end at C or D. Are there
also Eulerian paths starting at A or B? The answer is no. There are
still three roads to D; and, if we start at A or B, we can reach D by
one road, leave by a second, but can never get to travel the third road.
Thus, there are Eulerian paths starting and ending at C and D, but
none that starts or ends at A or B.
What is different about A and B that prevents Eulerian paths from
starting or ending there? Maybe we should look at it from the other
point of view: What is different about C and D that forces Eulerian
paths to begin and end there? Clearly, the fact that C is a dead end
is important. Once we reach C, we cannot leave; so, unless we start
at C, we must end there. But, what about D? The fact that there are
three roads to D seems to have played a role in our argument. If we
do not start at D, then, possibly, we can reach D by one road, leave
by a second and, if we are able to travel the third road to D, this
third road leaves us stranded at D. Thus, if we do not start at D, we
must end there.

180
CHAPTER 6
FIGURE 6.14
Can we generalize? We first consider one more example:
Playing around with Figure 6.14, we find several Eulerian paths
(for example, ABDCADEFD and ADEFDBACD), all of which seem
to begin and end at A and D.
What is special about A and D?
To help answer this question, we make some definitions.
ODD AND EVEN VERTICES
If the edge e connects the vertices U and W, then e is said to be
incident to U and to W.
The degree of a vertex U is the number of edges that are incident
to U. Each loop through U is counted twice since it has two ends at U.
If the degree of U is odd, then U is said to be an odd vertex;
otherwise, U is an even vertex.
For example, in Figure 6.14, A and D are odd vertices (of degrees
three and five respectively) and B, C, E, and F are even (each of degree
two). In Figure 6.13c, C and D are odd (of degrees one and three
respectively), and A and B are even. In Figure 6.13b, A, B, C, and D
are all odd—A, B, and C are of degree one, D is of degree three.
PRACTICE
PROBLEMS
6.A
1. [a] In each of the graphs in Figure 6.15, find the degree of each
vertex. A
A
FIGURE 6.15 (a) (b)
2. In each of the graphs in Figure 6.16, find the degree of each vertex.
A ABC
FIGURE 6.16
(a)

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY 181
We can now answer the question: What is special about the vertices
A and D in Figure 6.14, and the vertices C and D in Figure 6.13c?
The answer: They are odd, and all other vertices in these figures are
even. The following theorem applies in general.
Theorem 6.1 If a graph, ^, containing an odd vertex, U, has an
Eulerian path, then each Eulerian path for ^ must either begin or
end at U.
To prove this, we use an argument very similar to the one used
above to show that every Eulerian path in the graph of Figure 6.13c
must either begin or end at D. Namely, if we do not begin at U, then
each time we visit and leave U we use up two edges, one to reach U
and the other to leave U. Because each edge must be used exactly
once, and because there are an odd number of edges incident to U,
the edges incident to U must be used two at a time until, eventually,
we have left U and only one unused edge incident to U remains. When
we use this final edge incident to U, we will be stranded at U, and
so our path must end there.
Since a path can start at only one vertex and end at only one
vertex. Theorem 6.1 implies the following corollaries:
Corollary 6.1 If a graph, *^, containing two odd vertices U and V,
has an Eulerian path, then each such path must start at U and end
at V or vice versa.
Corollary 6.2 A graph that has more than two odd vertices cannot
have an Eulerian path.
Since the graph in Figure 6.13b has four odd vertices, it can have no
Eulerian path.
We are now able to answer Problems 6.1(a) and 6.1(b)
(page 174). If you have not yet solved them, try them again.
The graph of the problem. Figure 6.5a, has four odd vertices (B is Solution of
of degree five, and A, C, and D are of degree three); hence it can have Prohl^rn^ ^ iTo^
no Eulerian path. That is, a person cannot plan a walk as required. , . t/i \ *
On the other hand. Figure 6.5b—the graph of Problem 6.1(b)—has 0110 O. I \P)
only two odd vertices, C and D. Thus, it is still conceivable that an
Eulerian path for this graph exists. However, because C and D are of
odd degree, such a path—if one exists—must begin and end at C and D.
Thus, a walk starting at B cannot be planned, and neither can a walk
that terminates at its starting point.

182
CHAPTER 6
To plan a walk that starts at D is not very difficult; it is merely
a matter of trial and error. For example, one such path in Figure 6.5b is
DdBCABeDC.
This completes Problem 6.1(b).
Corollary 6.2 tells us about graphs with more than two odd vertices.
But what about graphs having two or fewer? Is it always possible to
find an Eulerian path for a graph having no more than two odd
vertices? Because graphs containing only even vertices are included
among those having no more than two odd vertices, let us consider
what happens when an Eulerian path begins at an even vertex. By an
argument very similar to that used to prove Theorem 6.1, we can prove
the following theorem.
Theorem 6.2 If a graph, ^, has an Eulerian path that begins at
an even vertex, U, then the path must also end at U; and, hence, the
path is an Eulerian circuit.
Proof Suppose the path begins at U and continues to W via an edge e.
Then the remainder of the path gives an Eulerian path, starting at W,
for the graph ^' obtained by omitting the edge e from the graph ^.
(For example, see Figure 6.17.) But, in ^', U is of odd degree (one
edge incident to U in ^ has been omitted); hence, every Eulerian
path for ^ ' that starts at W (does not start at U) must end at U.
Therefore, the Eulerian path for ^\ and hence for ^, ends at U.
(^'
w
A
Y
FIGURE 6.17
We are now ready to prove that graphs having no more than two odd
vertices always do have Eulerian paths. We begin with the following
theorem:
Theorem 6,3 If a connected graph, ^, has no odd vertices, then ^
has an Eulerian circuit.
The proof of this theorem actually gives a method, or algorithm, for
finding an Eulerian circuit. We outline the proof on the left below, and
on the right we apply the method to find an Eulerian circuit for the
graph in Figure 6.18a.

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
183
Begin at any vertex you like,
say, U. Start tracing a path and,
without retracing any edges,
continue until you get stuck.
Where will you be when this
happens? Each time you enter
and leave a vertex, you use up
two edges at that vertex, so that,
since all vertices are even, each
time you enter a vertex other
than U, there must remain at
least one edge by which to leave
that vertex. Thus, you cannot
get stuck at a vertex other than
U, and so you must be stuck at
U. That is, you have completed
a circuit F, beginning and
ending at U.
If r is an Eulerian circuit, you
are done. If not, then—since C-^
is connected—there must be at
least one vertex, W on F, for
which not all edges incident to
W have been used. (If e is an
unused edge incident to a vertex
R not on F, then—since C*^ is
connected—there must be a path
from U to R. The first place in
which this path uses an edge not
on F yields the desired vertex
W.) Starting with W, take a side
excursion, not using any edges
used in F, until you get stuck.
Since F uses an even number of
edges at each vertex, the number
of unused edges at each vertex
must still be even; and so, by
the argument above, the only
place you can get stuck is at W.
But then you can create a larger
circuit F', starting at U, as
follows: Follow F until you
reach W; then take the side
excursion, returning to W; then
continue on F back to U. If F'
Start at A:
r=ABCA
We are now stuck at A.
Choose a vertex on the circuit
for which unused edges remain.
For example, choose B and take
a side excursion (Figure 6.18c).
A
A BDEB
/ \
r' = ABDEBCA
(c)
Inserting this path, BDEB, into
the original circuit at B, we get
a new circuit F' = ABDEBCA
(Figure 6.18d).
(d)
FIGURE 6.18

184
CHAPTER 6
FIGURE 6.18
is not the desired Eulerian
circuit, you can continue as
above until the desired Eulerian
circuit is found. (Since the
number of edges is finite, the
process must terminate.)
(0
r"=ABDEFCEBCA
This is still not the desired
Eulerian circuit, so we take
another side excursion, say at E
(see Figure 6.18e).
A
A
/ \
/ \
EFCE
/ \
D E F
(e)
Now, inserting this path into F'
at E, we have the Eulerian
circuit shown in Figure 6.18f
on the left:
ABDEFCEBCA
PRACTICE 1. [a] Find an Eulerian circuit (if one exists) for each of the graphs
PROBLEMS
6.B
in Figure 6.19.
T yt 1
m •- •
Df
H I
(b)
G H
(c)
2. Find an Eulerian circuit (if one exists) for each of the graphs in
Figure 6.20.
A ABC
ABC
m
(a)
FIGURE 6.20

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
185
As a simple consequence of Theorem 6.3, we have the following
theorem:
Theorem 6,4 If a connected graph ^ has exactly two odd vertices,
U and W, then there is an Eulerian path for ^, starting at U and
ending at W.
Proof Add a new edge e connecting U to W. (We may have to leave
the plane to do this, but it always can be done.) This results in a new
graph ^' containing only even vertices. By the proof of Theorem 6.3,
^' has an Eulerian circuit F, starting by traveling via e from W to U
and then eventually returning to W. Omitting e from F gives the
desired Eulerian path for ^.
1. Find an Eulerian path (if one exists) for each of the graphs in PRACTICE
B
ABC
• •—
4- JF Dl + f F
i-
G
H I
(b)
H I
(c)
2. Find an Eulerian path (if one exists) for each of the graphs in
Figure 6.22.
PROBLEMS
6.C
T y1^[ T
[/ |e M
1x T 7t
* <lr '—i
H I
(b)
A B
F \^G
H
(c)
Theorems 6.3 and 6.4 take care of graphs with either zero or two
odd vertices. All that remains to consider, then, are graphs having
only one odd vertex. Actually, however, there are no such graphs. This
follows from the following theorem.

186
CHAPTER 6
Theorem 6.5 Every graph has an even number of vertices of odd
degree. (And hence there cannot be only one odd vertex.)
Proof Observe that if »S is the sum of the degrees of all the
vertices, and if E is the number of edges, then »S = 2£, since each
edge contributes exactly two in the count of S—onc at each vertex to
which the edge is incident. Thus, »S is always even. For example, the
graph in Figure 6.12a (page 178) has ten edges and six vertices—
A, B, C, D, E, and F—of degrees 2, 4, 3, 3, 4, and 4 respectively.
Note that
5 = 2 + 4 + 3 + 3 + 4 + 4 = 20 = 2- 10 = 2£.
Similarly, if Se is the sum of the degrees of all the even vertices,
then Se is a sum of even numbers and hence is even.
Finally, if So is the sum of the degrees of all odd vertices, then
S = Se + So. Thus So = S — Se) which is even. But So is a sum of odd
numbers. The only way a sum of odd numbers could be even is if
there are an even number of them. Hence, there must be an even
number of odd vertices.
Corollary 6.2 and Theorems 6.3, 6.4, and 6.5 tell us the whole
story about Eulerian paths. Summing up, we have the following:
1. A connected graph has an Eulerian circuit if and only if all vertices
are even.
2. A connected graph has an Eulerian path that is not a circuit if and
only if the graph contains exactly two odd vertices.
We are now able to complete Problem 6.1(c) (page 174). If you
have not already solved it, try again now.
^Oll jtion of Because part (c) of the problem requires us to find an Eulerian circuit,
_^ , I z i/ \ ^^ must construct the new bridge so that all vertices are even. As C
r rOOI©nn O. I\C^ and D are the only odd vertices in Figure 6.5b, adding a new edge
that joins them will make all vertices even. Thus, the new bridge must
be constructed from C to D (see Figure 6.23).
new bridge
'C^=]c:^«^,
D
(a)
FIGURE 6.23
(b)
f^
new bridf

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
187
PRACTICE
1. Plan a walk for a person starting at A in Figure 6.23, so that he PROBLEMS
or she can cross each bridge exactly once and end up again at A. A P)
^ MORE THAN TWO ODD VERTICES
The problem of finding an Eulerian path for a graph is sometimes
referred to as the "highway inspector's problem," because we try to
inspect every road once without unnecessarily retraveling any road.
Corollary 6.2 tells us that one inspector cannot accomplish this
task if there are more than two odd vertices; but several inspectors might
be able to succeed without duplicating any of the work. In fact, having
a different inspector for each stretch of road would certainly succeed;
but then more people would be hired than are really needed.
We now turn our attention to the question: How many inspectors
really are needed? That is, given a connected graph ^ with more than
two odd vertices, what is the smallest number of edge-disjoint paths
by which the graph can be covered? Edge-disjoint paths are paths
that have no edges in common with each other. For example, the
graph in Figure 6.24a has four odd vertices (B, C, D, and G) so it
has no Eulerian path. It can be covered, though, by two edge-disjoint
paths, DABEDGECB and CFIHFEHG (see Figure 6.24b).
• • H»
1 EJ/ 1
0Z
G H I
(a)
FIGURE 6.24
G
H
(b)
By Theorem 6.5, we may assume that the graph has 2k odd
vertices. Then the answer is given in the following theorem.
Theorem 6.6 A connected graph, C^, having 2k odd vertices may
be covered by k edge-disjoint paths.
Proof From our previous discussions, it is clear that fewer than k
paths will not work, because each path can remove the oddness of at
most two vertices. On the other hand, k paths always suffice.

188
CHAPTER 6
On the left below, we show that k paths are always enough, while
we simultaneously consider an example on the right.
We begin by pairing off the
odd vertices. Say they are
u„ w„u,, w,,...,u,, w,.
We now form a new graph '^'
by first adding k new vertices,
Ti, T2, ..., Tk and then adding
2k new edges e^, e2, ..., e/;,
fi, f2, ..., f/t, where, for each i,
e, connects U, and T,, and f,
connects W, and T,. (Again, we
may have to leave the plane to
do this.)
Since one edge has been added
to each of the odd vertices of C-^,
and all new vertices are of degree
two, C-^' has only even vertices.
Hence, ^,^' has an Eulerian
circuit. Since T, can be reached
only from U, or W,, each of the
segments e, T, f, (or f, T, e,) must
appear in the circuit.
Removing these cuts the
circuit into k pieces—the k paths
required to solve the original
problem.
The graph in Figure 6.25a has
four odd vertices. We label these
U„ W„ U2, and W2 (and label
the other vertices A, B, C, D,
and E), to get Figure 6.25b.
A U, B
T T T
1 IE 1
T T T
i i i
w,
(a)
/
IB
W,
/ M
c
W2
(b)
(c)
We obtain an Eulerian circuit
for Figure 6.25c, starting with
UjeiTifiWi. For example,
U,e,T,f, W,CW2 EU2 DW2 f2 T2-
e2U2AU,EW,BU, (Figure 6.25d).
"Mm
(d)
Omitting the segments e,Tifi
and f2 T2 ^2 J we obtain the two
paths W1CW2EU2DW2 and
U2AU,EW,BU„ which give
Figure 6.25e, an edge-disjoint
covering of the graph in
Figure 6.25a.
£33
CO
FIGURE 6.25
(e)

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
189
Instead of hiring several highway inspectors to cover a given network,
we may allow a highway inspector to retrace some roads, if necessary.
We can now ask the following question: How can one highway
inspector cover the entire network while keeping the number of
retraced edges to a minimum?
Retracing an edge is essentially the same as adding a new edge that
is incident to the same two vertices as the edge that is to be retraced.
Our aim is to add these new edges so that all but two of the vertices
become even and therefore an Eulerian path exists.
For example, to cover the graph in Figure 6.26 with a minimal
amount of retracing, we must eliminate two odd vertices.
We do this by adding an edge (or retracing an edge) between two
of the odd vertices, say, Uj and W^. Actually, we must retrace two
edges, for example, UjEWi or UiEWj. We then get an Eulerian path,
such as
U, AU,BW,EU,BW,CW2 DU^ EW,
from U, to W2.
If we want to start at a particular point and end at that same point,
then we need an Eulerian circuit and hence must eliminate all odd
vertices. In the above example, this can be done by retracing two
more edges (U2 DW2, for example).
U, B
• • •
1 JE 1
T T T
i i i
D W2 (
FIGURE 6.26
1. [a] Completely cover each of the graphs in Figure 6.27 by the
fewest possible edge-disjoint paths.
ABC
T^^
|/T\
^\57T
H
K L M
(a) (b)
FIGURE 6.27
2. Completely cover each of the graphs in Figure 6.28 by the fewest
possible edge-disjoint paths.
A B
A B
cAA.
w
F G
(a)
FIGURE 6.28
^ \ 1
I
D
i
\e/
(b)
F \
;
PRACTICE
PROBLEMS
6.E

190
CHAPTER 6
3. [a] Trace each of the figures in Problem 1 above without lifting
your pencil from the page, in such a way that you retrace as few
edges as possible.
4. Trace each of the figures in Problem 2 above without lifting your
pencil from the page, in such a way that you retrace as few edges
as possible.
5. [a] Starting at A and ending at A, trace each of the figures in
Problem 1 so that you retrace as few paths as possible.
6. Starting at A and ending at A, trace each of the figures in Problem 2
so that you retrace as few of the paths as possible.
D " C
FIGURE 6.29
i> DIRECTED GRAPHS
Another interesting variation of the Eulerian path problem occurs if
we assume that each edge can be traveled in only one direction, like a
one-way street. The resulting geometric structure is called a directed
graph or digraph.
As before, we want to know when we can traverse every edge of the
graph exactly once; but this time we must obey the "one-way"
signs. Although this new problem seems more difficult, its solution is
actually similar to that of the original problem.
We define the in-degree of a vertex as the number of edges incident
to that vertex which are directed toward the vertex. We define the
out-degree of a vertex analogously—the number of incident edges
which are directed away from the vertex. For example, in Figure 6.29,
the in-degrees and out-degrees are:
in-degree
out-degree
A
1
1
B
2
2
C
1
2
D
1
2
E
1
3
F
4
0
We now have the following theorem.
Theorem 6,7 A connected digraph '// has an Eulerian dicircuit
(that is, an Eulerian circuit which obeys directional signs) if and only if
the in-degree of each vertex is equal to the out-degree. (Note that this
condition implies that every vertex is of even degree.)

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
191
We leave the details for you to supply; see how similar they are to
the discussion above concerning graphs in general.
1. By analogy, define what is meant by an Eulerian dipath.
2. [a] If possible, find an Eulerian dicircuit or Eulerian dipath for
each of the graphs in Figure 6.30. If it is not possible, explain why
not.
->--T
->—fF
G H
(a)
FIGURE 6.30
t—>-
A \\ A
H
(b)
(c)
PRACTICE
PROBLEMS
6.F
3. If possible, find an Eulerian dicircuit or Eulerian dipath for each of
the graphs in Figure 6.31. If it is not possible, explain why not.
FIGURE 6.31
(b)
Before ending this discussion, we should point out that problems
involving Eulerian paths are occasionally disguised in various ways.
Problem 6.2 (page 174) is such a problem. Try it again now
if you have not already solved it.

192
CHAPTER 6
Solution of If we represent the rooms and exterior by letters, then we see that
DrQ^vl^ppi A p the problem really requires us to find an Eulerian path on the graph
in Figure 6.32b.
c
C
1 1
A
d
1
B
e f
D
g
h ^ '
1 1
(a)
FIGURE 6.32
Because this graph has two odd vertices, B and E, there exist
Eulerian paths beginning at B and ending at E or vice versa. However,
it is not possible to find such a path that begins and ends at F, the
exterior of the house.
Thus, the answer to Problem 6.2 is "No. It is not possible."
HAMILTONIAN CIRCUITS
Problem 6.3 (page 174) is a two-dimensional representation of a game
called "Around the World," which was invented by Sir William Rowan
Hamilton in 1857. The original game consisted of a solid regular
dodecahedron (one of the five regular solids) with each of the 20
corners labeled with the name of a city—see Figure 6.33b. The object
of the game was to travel around the world, along the edges of the
dodecahedron, visiting each of the 20 cities exactly once.
Montreal
Glasgow
Boston \Amsterdam
Berlin
Rome
FIGURE 6.33

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
193
Although Problem 6.3 seems to be closely related to our discussion
of Eulerian circuits, it is actually quite different. With Eulerian circuits,
each edge has to be traversed exactly once; here, some edges need
not be traversed at all. With Eulerian circuits, we could visit a vertex
as often as necessary (as long as we use different roads each time to
reach and leave the vertex); here, each vertex must be visited exactly
once.
Given a graph, a path that visits each vertex exactly once is called
a Hamiltonian path; and, if it returns to its starting point, it is called
a Hamiltonian circuit.
As with Eulerian paths, a graph must be connected in order for a
Hamiltonian path to exist; but the similarity ends there. Although
there are some theorems that give conditions under which Hamiltonian
paths will exist, there are no known criteria that completely characterize
those graphs which have Hamiltonian paths; nor is there any simple
algorithm that enables us to find a Hamiltonian path if one exists.
Each graph must be examined individually, and finding a Hamiltonian
path—if one exists—is a matter of luck, insight, and reasoned trial
and error.
If you could not solve Problem 6.3 (page 174) before,
try it again now.
With Problem 6.3, finding a Hamiltonian path is not too difficult. With
a little experimentation, we find, in fact, a Hamiltonian circuit:
ABCDENMLKJIHGQRSTPOFA
There are many other circuits that also work.
Solution of
Problem 6.3
Although there is no simple algorithm that tells us how to find a
Hamiltonian path if one exists, two observations are worth noting:
1. If there is a vertex of degree one, then any Hamiltonian path must
start or end there.
2. If there is a vertex of degree two, then either the path must start
or end there, or else both edges incident to the vertex must be
traveled.
These observations will be helpful in attacking some of the exercises
at the end of the chapter. An example of how they can be applied
is given in the discussion that follows.

194
CHAPTER 6
THE KNIGHT'S TOUR
As with Eulerian path problems, Hamiltonian circuit problems often
appear in disguise. One well-known example is the knight's tour of
the checkerboard.
The problem may be stated as follows:
Place a knight somewhere on the checkerboard; and then move it
so that it visits each square of the board exactly once.
If the square from which the knight started can be reached in one
knight's move from the final square, then the tour is said to be
reentrant.
In order to discuss the knight's tour problem, we must first be
familiar with the knight's move in chess. We may describe this move
as follows: A knight moves horizontally or vertically two spaces and
then turns left or right for one space (an "L" in essence, as Figure
6.34a shows). Thus the knight—indicated by N in Figure 6.34b—could,
on its next move, move to any of the squares indicated by an X.
~l
4-
\M^
@iE]
(a)
X
X
X
X
N
X
X
X
X
(b)
FIGURE 6.34

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
195
An example of a reentrant knight's tour on a 5 x 6 checkerboard is
illustrated in Figure 6.35. The knight starts in the upper lefthand
corner, then moves to the square labeled 2, then 3, and so on.
But what does this have to do with graphs? We can think of the
squares of the checkerboard as the vertices of a graph, with edges
connecting two vertices if and only if a knight could move directly from
one to the other. For example, for the 3x4 case, we get the graph
shown in Figure 6.36. Note that only the 12 labeled vertices are
considered as points of intersection of the edges.
1
28
21
10
3
22
11
2
29
20
27
30
23
4
9
12
15
8
19
24
7
26
17
14
5
16
13
6
25
18
FIGURE 6.35
A
E
I
B
F
J
C
G
K
D
H
L
(a)
FIGURE 6.36
The knight's tour requires that he visit each vertex exactly once—
that is, we must find a Hamiltonian path.
For the graph in Figure 6.36, such a path can be found fairly easily
as a result of our observation pertaining to vertices of degree two.
If a vertex is of degree two, then both edges incident to that vertex
must be traveled in any Hamiltonian circuit or in any Hamiltonian
path which does not start or end at that vertex. Applying this to
Figure 6.36, we see that vertices A, G, H, and I are of degree two
(as are D, E, F, and L). Hence, a Hamiltonian path must contain all
of the edges shown in Figure 6.37, with the possible exception of one
incident to the starting or ending point of the path.
Similarly, the other six vertices give rise to the edges pictured in
Figure 6.38. The only connections between these two circuits are the
edges B-K and C-J. Thus, to find a Hamiltonian path, we must
traverse the first six vertices (A, B, G, H, I, and J) ending on B or J,
and then we must continue with a traversal of the other six vertices,
beginning at K or C. One path that works is:
AGIBHJCEKDFL
Others may be found similarly.
There is no Hamiltonian circuit for the Figure 6.36 graph—that is,
a reentrant knight's tour on a 3 x 4 checkerboard is not possible.
With larger boards, graph theory can still be used to represent the
FIGURE 6.37
FIGURE 6.38

196
CHAPTER 6
knight's tour problem. But the lack of a simple algorithm for finding
Hamiltonian paths often causes the graph-theoretical representation to
be of only limited value.
OTHER APPLICATIONS
So far, we have considered problems in which the object was to travel
along all edges or through all vertices of a graph. Frequently, however,
we are satisfied with much less. Suppose, for instance, that there is
a system which may be in any of several "states" (conditions,
situations), and the object is to get from one particular state to another
one. Then the problem may be represented graphically as follows: We
consider the possible states of the system as the vertices of a graph,
and connect two states by an edge if it is possible to go directly from
one state to the other. The problem then becomes one of finding any
path between the vertices corresponding to the particular states in
question.
" Crossing problems " are a class for which this approach is sometimes
helpful. Problem 6.4 is a classical example.
If you have not already solved Problem 6.4 (page 175),
try it again now.
Solution of
Problem 6.4
In Problem 6.4, let M represent the missionary who can paddle, C the
cannibal who can paddle, m, and ni2 the other missionaries, and c^
and C2 the other cannibals. Then, according to the conditions of the
problem, the only sets of people who can be on the original side of
the river together with the canoe are:
Mmim2Cc,C2, Mm,m2Cc,, Mmim2CiC2, Mmjm^C, Mm,m, c,,
Mm, Cc^, Mm, C1C2, mim2 Cc,, Mmjm,,
Cc,C2, Cc,, MC, Mc,, m, C, or C,
(where / = 1 or 2, and 7 = 1 or 2). All other states are prohibited,
either because cannibals would outnumber missionaries on one of the
sides of the river, or because there is no way the boat could have
returned (no one on the side with the boat can paddle).
We represent the possible states as the vertices of a graph, and
connect two vertices by an edge if it is possible to go directly from one
of the states or situations to the other. Note that we consider only
states with the canoe at the original bank, so a "move" consists of
someone paddling over and someone paddling back. For example, we
can go from the state Mmim2C to the state MmjCc^ by having M

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
197
row m2 over and bring C2 back. The total graph is indicated in Figure
6.39. Note that, in getting from one state to the next, we must consider
Mm.m^c,
Mc,
start
Mm,m2Cc,C2
FIGURE 6.39
finish
Mm,m2C,c
what the situation would be when the canoe is brought to the second
side of the river. Thus, for example, from Mmim2 C1C2, we cannot have
Mm, paddle over, for that would leave two cannibals and one
missionary on the near shore; nor can we have Mc, paddle over,
because, when they arrive, the cannibals on the second shore will
outnumber the missionary.
From the graph, it is now easy to find the solution to the problem.
All that is needed is a path from start to finish. It is not only easy to
find such a path but, in fact, we can find one that minimizes the
distance to be paddled. We exhibit the solution in a table (Figure 6.40).
in the boat
start
trip 1. out Mc,
return M
trip 2. out Cc2
return C
trip 3. out Mm,
return Mc2
trip 4. out MC
return Mc,
trip 5. out Mrn2
return C
trip 6. out Cc,
return C
trip 7. Cc2
finish
on the near shore
Mm, m2Cc,C2
m, m 2 Cc 2
m jm2Cc2
Mm,m2
Mm,m2
m2C
m2C
m2C2
C,C2
C1C2
C2
C2
on the far shore
Cl
C,C2
C1C2
m,c,
m,c,
m,C
m,C
Mm,m2
Mm,m2
Mm,m2C,
Mm,m2C,
Mm, m2Cc,C2
FIGURE 6.40

198 CHAPTER 6
The same general approach used in Problem 6.4 may also be used in
many other types of problems. The main difficulty is that, if the
number of possible states is large, the graph will have many vertices
and will become time consuming to construct and difficult to read.
K> COLORING GRAPHS AND MAPS
No chapter on modern graph theory would be complete without
mention of the recent proof of the Four Color Conjecture, which
deals with the coloring of maps in the plane. According to the
evidence available, in 1852 Francis Guthrie first raised the following
question*: Given any map, can it be colored in at most four colors so
that countries that share a common border are colored in different
colors? Guthrie conjectured that the answer was yes. What makes this
conjecture so noteworthy is that, although many of the best
mathematical minds of the nineteenth and twentieth century worked
very hard to try to prove (or disprove) the conjecture, the problem
remained unsolved until 1976, when Kenneth Appel and Wolfgang
Haken completed a proof. We can now refer to the result as the Four
Color Theorem.
The theorem may be stated very simply as follows:
Theorem 6,8(The Four Color Theorem) Every planar map may
be colored in at most four distinct colors, so that no two regions
which share a common border have the same color.
The statement of the theorem requires some explanation. Any planar
graph divides the plane into a number of regions. For example, the
graph in Figure 6.41 divides the plane into three regions. (Note that
what might be called the exterior of the graph is counted as a region—
region 3.)
The graph together with these regions is called a planar map,
provided that each edge of the graph is on the boundary of two
distinct regions.
Two regions of the map are said to share a common border if some
edge of the graph is part of the boundary of each region. Two
regions of a map that meet only in one point or a finite number of
points are not considered to share a common border. For example, in
Figure 6.42, region 3 shares a common border with 2, 4, and 5, but
does not share a common border with 1. Similarly, 2 shares a common
border with 1 and 3 but does not share one with 4 or 5.
* See [2] and [7] pp. 90-93 for fuller discussions of the history of the conjecture.

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
199
FIGURE 6.41
FIGURE 6.42
It should be poinred our rhar a region or counrry of the map must
be connected. That is, we would have to consider the United States
and Alaska as different regions because they are separated by Canada.
These two regions need not receive the same color.
The proof of the Four Color Theorem is lengthy and beyond the
scope of this book. It depends on an involved case analysis that is
carried out with the help of computers.
On the other hand, it is interesting to note that a weaker Five
Color Theorem was proved in 1870, over a century ago. This theorem
is as follows:
Theorem 6,9 (The Five Color Theorem) Any planar map may be
colored in at most five distinct colors so that no two countries that
share a common border have the same color.
Of course, now that the Four Color Theorem has been proved, the
Five Color Theorem is no longer of great interest. However, it does
suggest an interesting question: Might it not be possible to improve
the Four Color Theorem to a Three Color Theorem? In fact, it is nor
possible to do so, as the map in Figure 6.43 shows.
In this map the four numbered regions all touch each other, so at
least four colors are required to color this map.
On the other hand, there are some maps that may be colored in
fewer than four—or even three—colors. For example, we have the
following theorem:
Theorem 6.10 A planar map may be colored in two colors if and
only if each vertex is of even degree.
Given any planar map, if we shrink each region to a point or vertex
and consider each edge between two regions as a bridge or edge
connecting the vertices (as we did in the Konigsberg Bridge Problem),
then we get a new graph called the dual of the original map. For
example, the graph in Figure 6.44b is the dual of the map in 6.44a.
The vertices of this graph are the regions of the map.
FIGURE 6.43

200
CHAPTER 6
111
(a)
FIGURE 6.44
(b)
If the original map is colored, then the vertices of the graph we
obtain will also be colored. In fact, no two different vertices that are
connected by an edge will have the same color.
This notion may be applied to define a coloring for any graph—
color the vertices so that no two distinct vertices that are connected
by an edge receive the same color. It may be shown that theorems about
map coloring can be translated into theorems about coloring graphs.
In particular, any planar graph may be colored (in the above sense)
in at most four colors.
Instead of coloring the vertices of a graph, we may color the edges
so that no two edges incident to the same vertex receive the same
color. With this notion of coloring, there can be no analog of the
Four Color Theorem, because a vertex of degree n will necessitate at
least n colors (see Figure 6.45).
FIGURE 6.45
PRACTICE
PROBLEMS
6.G
1. [a] Excluding the exterior regions, what is the fewest number of
colors needed for each of the maps in Figure 6.46 so that no two
regions which share a common border receive the same color?
(b)
FIGURE 6.46

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
201
(c)
(d)
FIGURE 6.46
2. Including the exterior regions, what is the fewest possible colors
needed for each of the maps in Figure 6.46 so that no two regions
which share a common border receive the same color?
3. [a] If you were to color the edges of each of the graphs in Figure 6.47
so that no two edges which are incident to the same vertex receive
the same color, what is the minimum number of colors you need?
T 1 T ?
T T T T
T T T T
i 1 i i
(a)
FIGURE 6.47
(b)
(c)
THE CHAPTER IN RETROSPECT
This chapter has presented a modern topic in mathematics which is
different in flavor from the basically algebraic and arithmetic topics
of Chapters 3, 4, and 5.
The chapter has also pointed out how a mathematical theory can
be developed from an apparently simple question: Is it possible to
take a certain walk in Konigsberg without retracing any part of the trip ?
By abstracting the basic elements of the problem and defining certain
terms—graph, vertex, edge, path, and so on—we can restate the
question simply as, "Does the graph representing the problem have
an Eulerian path or circuit?"

202
CHAPTER 6
Our next step was to try a few problems, and we discovered that the
parity of the vertices was somehow involved. These discoveries led to
the development of some conjectures, which became theorems when
we finally proved them. The proofs were based on definitions,
previously proved theorems, and logical reasoning.
The theorems we finally proved enabled us to answer not only the
original question, but also related ones.
Much mathematical theory develops along the above lines, although
many years may pass between conjecture and proof. For example, as we
saw above, it took over a hundred years for the Four Color Theorem
finally to be proved. And, as we saw in Chapter 4, Fermat's famous
conjecture is still not proved today.
Once a mathematical theory has been developed in response to a
question, other applications frequently appear. For example, the
Konigsberg problem is of interest in terms of highway inspection and
its analogs—police patroling streets, trash collection, and others.
The question is relevant and practical, even though it initially arose
in a recreational setting.
The exercises at the end of this chapter can all be stated or solved
graph-theoretically. However, in some cases, the graph-theoretic
approach is not practical without the aid of a computer—there are too
many possibilities to be handled efficiently. Nevertheless, we include
these problems so that you will be aware of their graph-theoretical
connection.
Exercises
Eulerian Paths and Circuits
6.1. E. Sterner of Philadelphia,
Pennsylvania, is planning her first trip through the
part of the United States pictured in Figure
6.48. She intends to fly to Omaha, Nebraska,
beginning her tour there. The remainder of
her trip is to be made by car, along such a
route that she crosses the border between each
pair of neighboring states exactly once. (That
is, she crosses the Nebraska-Wyoming border.
the Nebraska-South Dakota border, the
Wyoming-South Dakota border, etc.)
Can you help Ms. Sterner plan her trip?
FIGURE 6.48
Mont.
Wyo.
N.D. \
_[ 's
1 s-^- i
j Neb. ^
-i \

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
203
6.2. [a] Ms. Sterner (see Exercise 6.1) decided
that she has enough time to add four more
states to her itinerary (see Figure 6.49). She
still wants to cross the border between each
pair of neighboring states exactly once.
FIGURE 6.49
If she flies to Omaha, Nebraska, in what
state must she catch a return flight in order
for her to be able to complete such a trip?
6.3. [h] Gia Graphy has a map showing a
cluster of states in the United States. She has
labeled each state with a number equal to the
number of states bordering it on the map.
Prove that the number of states which have
been labeled with odd numbers is even.
6.4. [h] Six famous numismatists, Messrs.
Dollar, Pound, Frank, Mark, Lira, and Yen,
met to trade coins with each other. Each trade
that took place was between only two people.
After the meeting, each of the six was asked
how many people he had traded with. The
answers were 5, 4, 2, 1, 3, and 2
respectively.
Prove that at least one person is mistaken.
6.5. [h] [a] Two cells of a chessboard are said
to be separated by a knight's move if it is
possible for a knight to move from one of
them to the other in one move.
(a) Is it possible for a knight to travel over
a 4 X 4 chessboard in such a way that it
moves exactly once between each pair of cells
which are separated by a knight's move? (If it
moves from cell X to cell Y, then it cannot also
move from cell Y to cell X.)
(b) For what values of n is it possible for a
knight to make such a trip on an w x w board?
(c) Considering a move and its reversal as
two different moves, plan a trip for a knight
starting on cell A of the 3x4 board pictured
in Figure 6.50 so that every possible knight's
move is made exactly once.
A
E
I
B
F
J
C
G
K
D
H
L
FIGURE 6.50
6.6. [h] (a) If possible, plan a trip for a bishop
on the 7x7 chessboard in Figure 6.51 so that

204
CHAPTER 6
it moves exactly once between each diagonally
adjacent pair of black cells. If such a trip is not
possible, explain why not.
(b) If possible, plan such a trip on the
white cells. If no such trip is possible, explain
why not. (Use cordinate notation to represent
the cells. For example, G-1 is the lower right
hand corner, A-7 is the upper left hand
corner, D-4 is the center square.)
A B C D
FIGURE 6.51
F G
6.7. [s] [a] Figure 6.52 shows the layout of
Woodlawn Gardens. Each line represents a
path along which flowers are growing, and
each path is exactly 100 meters long. The
entrance to the gardens is at the point labeled
A, and the exit is at point B.
FIGURE 6.52
Mr. E. Ficiency was hired to conduct a tour
of the entire gardens in such a way that the
necessary amount of walking would be held to
a minimum.
What route should he follow? What is the
total distance walked?
6.8. [h] Maggie Zehan is working her way
through college by selling subscriptions to
periodicals. Today, she is planning to try her
luck at a housing development, a map of
which is shown in Figure 6.53. Each street has
houses on only one side. Maggie wants to
cover every street, but she also wants to keep
her walking to a minimum.
A
200
r>
X^ 200
i
B
o
E
O
200
^
200 \l
i
G 200 H 200 I
(distances in meters)
FIGURE 6.53
(a) [a] What is the minimum distance she
will have to walk if she can start and end at
any points she chooses? What should her route
be?
(b) If the only parking lot in the
development is at point A (that is, she must start and
end at A), what is the minimum distance she
will have to walk? Describe her route.
6.9. The Ramada County Department of
Highways has just resurfaced the county roads,
and now the yellow stripe down the middle of
each road must be repainted. The truck used
for this purpose is very inefficient as far as
gas consumption is concerned, and so the
Department would like to have the truck
travel the shortest distance possible. A road
map of the county is shown in Figure 6.54.

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
205
W. Midville
E. MidviUe
S.W. Midville S.E. Midville
(distances in kilometers)
FIGURE 6.54
The county truck is garaged in Midville, and
it must return there when the job is done.
What route should it follow?
6.10. [h] The magician placed a standard
28-piece set of dominoes on the table,
removed the doubles (double zero through
double six) and turned his back. He then
asked the volunteer to remove any one of the
dominoes and to arrange all the remaining
dominoes in a chain, in accordance with the
rules of the game of dominoes (touching parts
of two different dominoes must both contain
the same number; no side branches are
allowed).
When the chain was completed, the
magician turned around, took a very quick
glance at the arrangement on the table, and
said, "The domino you removed is the two-
five."
How did he know? Explain why the trick
works.
M 6.11.1 [s] I [a] There are 27 different three-
digit numbers that can be made from the digits
1, 2, and 3 — 111, 121, 123, 312, etc.
Use graph theory to determine how to
place nine Ts, nine 2's, and nine 3's in a
circular arrangement so that each of the 27
numbers appears exactly once when all sets of
three consecutive digits around the circle are
read in a clockwise direction. That is, the
sequence shown in Figure 6.55 would give 121,
213, 132, etc. ([64], Problem 23c, p. 110)
FIGURE 6.55
6.12. [h] Place four O's and four I's in a
circular arrangement, so that each of the triples
(0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0), (0, 1, 1),
(1, 1, 0), (1, 0, 1), and (1, 1, 1) appears
exactly once when all sets of three consecutive
digits are read around the circle in a clockwise
direction.
Hamiltonian Circuits
6.13.I[h]|[a| a saleswoman, starting at town
A on the map in Figure 6.56, wishes to visit
every town on the map exactly once, returning
to A at the end of her trip.
FIGURE 6.56

206
CHAPTER 6
How many different possible routes could
she plan that would satisfy these conditions?
6.14. [h] a salesman wishes to visit all the
cities on the map in Figure 6.57 and to return
to his starting point.
What route should he follow if he does not
wish to visit any city more than once?
FIGURE 6.57
6.15. [h] Albatross Airlines serves the nine
cities on the map in Figure 6.58, with the
flights indicated (two points are connected by
an edge if and only if there is a direct Albatross
flight between the corresponding cities).
Show that a vacationer wanting to visit each
of these cities exactly once would not be able
to end her trip at her starting point if she
uses only Albatross Airlines for transportation.
G ^—^ H
FIGURE 6.58
6.16. [a] On the graph in Figure 6.59, start
at the vertex labeled start, and travel along the
graph so that you visit every vertex exactly
once, ending at a vertex labeled T. So far,
most people who have seen this puzzle have
said, "This is too difficult."
What do you say?
FIGURE 6.59
M 6.17. [s] [a] Each of the 22 vertices on the
graph in Figure 6.60 represents a shop that
serves as a front for a bookie. The edges
joining these vertices represent the only streets
connecting these shops. The police have placed
stakeouts outside each shop with orders to
arrest anyone passing the same shop twice.
Can a numbers runner make the rounds of
all the shops without being arrested? ([38],
Chapter 1, Problem 30)
FIGURE 6.60
6.18. [s] The lighting technician at the Foot-
light Theater has three colored filters (red,
green, and blue) that can be placed over the
main spotlight. By varying which filters are
on and off he can obtain eight different lighting
effects. The filters may be changed one at a
time either by adding a filter not being used
or by removing one that was just used.
Assuming that he starts and ends with no
filters, how can he test all eight lighting
effects, without repeating any effect except the
final one?

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
207
6.19. [a] (a) Is it possible for a rook to start
on a corner cell of a 6 x 6 chessboard and to
visit every cell of the board exactly once in
such a way that it ends on its starting cell?
(Note that a rook may move horizontally or
vertically but not diagonally. A cell is
considered visited if the rook passes over that cell
or lands there.)
(b) \h\ What happens on a 5 x 5
chessboard ?
(c) Generalize.
6.20. [h] Consider the 7x7 chessboard in
Figure 6.61. Recall that a bishop may move
diagonally but not horizontally or vertically.
(a) Describe a bishop's tour of the black
cells of the chessboard (each black cell is visited
once).
(b) Show that a bishop's tour of the white
cells of the board is not possible.
(c) Show that a bishop's tour is not possible
on the cells of either color of an « x w board,
where n is any even integer greater than two.
(d) Show that a reentrant bishop's tour (one
that ends where it starts) is not possible on the
cells of either color of an w x w board, where
n is any integer greater than five.
6.21. [h] Describe a knight's tour on the 3 x 7
chessboard pictured in Figure 6.62.
A
H
O
B
I
P
C
J
Q
D
K
R
E
L
S
F
M
T
G
N
U
FIGURE 6.62
-¥ 6.22. [h] Describe a reentrant knight's tour
on the 3 X 10 chessboard pictured in Figure
6.63.
1
11
21
2
12
22
3
13
23
4
14
24
5
15
25
6
16
26
7
17
27
8
18
28
9
19
29
10
20
30
FIGURE 6.63
6.23. [h] King Arthur has hired a consultant,
Sir Cumference, to plan a seating arrangement
for Arthur and nine of his knights to sit
around the round table. This would not be a
difficult task, were it not for the jealousies and
petty rivalries that exist among the knights.
FIGURE 6.61

208
CHAPTER 6
Arthur insists that Lancelot should sit on his
right and that Kay should occupy the seat on
Arthur's left; Bedivere refuses to sit next to
anyone but Lionel and Tristan; Gawain won't
sit next to Tristan, Lancelot, or Lionel; Gareth
won't sit next to Galahad, Lancelot, or Kay;
Perceval objects to sitting next to Galahad,
Lancelot, or Lionel; Tristan refuses to sit next
to Lancelot, Perceval or Kay; Galahad will sit
next to anyone except Gawain and Kay; and
Lionel will not sit next to Galahad. The other
two knights are not particular about whom
they sit next to.
Help Sir Cumference find a suitable seating
arrangement.
Crossing Problems
6.24. A farmer owns a wolf, a goat, and a
cabbage, and he wishes to transport them to
the other side of a river. Unfortunately, the
only available boat is large enough to hold
only the farmer and one of his possessions at
one time. (It is a giant cabbage!) The farmer
cannot afford to leave the wolf and the goat
together unchaperoned, for the former would
eat the latter. Similarly, the goat and cabbage
may also not be left together unattended.
How can the farmer safely transport all three
of his possessions across the river?
6.25. [h] Three married couples want to cross
a river. The only boat available is capable of
holding only two people at a time. This would
present no difficulty were it not for the fact that
the women are all very jealous, so that each
woman refuses to allow her husband to be in
the presence of another woman unless she
herself is also present.
How should they cross the river with the
least amount of rowing ?
6.26. [h] Five couples want to cross a river.
There is a canoe that holds three people. No
man will allow his wife to be in the presence
of another man unless he himself is present.
How can the crossing be made with the least
amount of rowing?
6.27. How can n couples cross a river in a
boat that holds four people, so that no man
is in the presence of a woman unless both his
wife and her husband are present?
The following problems are similar to the
previous ones and may be represented graph-
theoretically^ but the graph-theoretic approach
is not always a reasonable method of attacking
them.
-¥ 6.28. Four couples want to cross a river.
There is a kayak that will hold only two
people. No man will allow his wife to be in
the presence of another man unless he himself
is also present.
(a) [s] Show that the crossing is not
possible under the stated conditions.
(b) [a] Show how the crossing can be made
if there is an island in the middle of the river.
6.29. [a] Five men come to a river and wish
to cross. The only boat they can see is a very
small one which is being used by two boys.
In fact, the boat is so small that, although the
boys can both fit in it, it can hold only one
adult and cannot even accommodate one man
together with one child.
Enlisting the aid of the boys, what is the

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
209
smallest number of trips required for the men
to cross, with the boys ending up in the boat ?
6.30. [h] [a] Jon, Val, and John plan to steal
a treasure from a castle tower, with the aid of
John's trained dog, Hugo. The men will have
to escape through the tower window, by means
of a tackle consisting of a long rope with a
basket at each end (outside the window) and a
pulley. However, they must be careful. If the
difference in weight between the contents of
the two baskets is more than 20 lb, then the
heavier basket will descend too rapidly for the
safety of any of the robbers or of Hugo;
however, the treasure chest would not be
harmed by the shock of the impact.
John weighs 210 lb, Jon weighs 100 lb,
Val weighs 120 lb, and Hugo weighs 20 lb.
The treasure is in a chest weighing 60 lb,
to which the men will steal the key (which
weighs 1 lb).
How can the men escape from the tower
so that no one is hurt and no one is double
crossed? (The treasure chest is worthless
without the key.)
6.31. [h] When Jon, Val, and John (see
Exercise 6.30) actually carried out the theft,
everything went smoothly during the first part
of the operation. After they were all safely
down from the tower, they opened the chest
and each took a share of the treasure inside.
Each person placed his share in a bag he had
brought for the occasion. Since Jon
masterminded the entire operation, he received the
lion's share of the loot—$30,000 worth. John
received $20,000 worth, since his dog had been
vital to the success of the plan. Val received
the remaining treasure, which was worth
$10,000. On the way home from the castle, the
men came to a river. The only boat available
could hold only two men at a time; or it
could hold one man and the dog; or else, one
man and one bag of loot.
How can they cross the river so that no man
is ever alone (ignore Hugo) with more than his
share and no two men are ever alone with more
than their combined shares? (In case you were
wondering, Hugo, the trained dog, does not
know how to row.)
Miscellaneous Graph-Related
Problems
6.32. [s\ Show that if there are more books
in a library than there are pages in any one
book, then at least two of the books must
contain the same number of pages.
6.33. Show that in a room full of people
(more than one person), there are at least two
people who have the same number of friends
in the room. (Assume that if B is A's friend,
then A is also B's friend.)
6.34. [h] Place a marker on a dot in the
diagram in Figure 6.64, then slide it along

210
CHAPTER 6
FIGURE 6.64
one of the two lines from that dot to an
adjacent dot. (For example, place it on 7 and
slide it to either 2 or 3.) Leave the marker
on this new dot. Place a second marker on an
empty dot and slide it to a vacant adjacent
dot. Continue in this manner until you are no
longer able to do so. Your goal is to place eight
markers.
How should you do this?
6.35. [h] [a] Move the knights in the diagram
in Figure 6.65 so that the white knights
change places with the black knights.
How can this be accomplished in the fewest
moves?
WN
BN
WN
BN
FIGURE 6.65
6.36. [h] [a] Place a white marker on each of
the numbers 1, 2, and 3 in the Figure 6.66
diagram, and place a blue marker on 10, 11,
and 12. The object of this puzzle is to move
the markers until the white ones occupy the
spaces originally occupied by the blue ones,
and vice versa. A move consists of moving one
marker from one vertex of the graph to an
adjacent vertex (along an edge of the graph).
However, at no time may a white marker and a
blue marker occupy graphically adjacent
edges. That is, if a white marker is on 4, then
no blue marker may be on 3, 9, or 11.
10 11
FIGURE 6.66
What is the fewest moves in which the
transfer may be made?
6.37. [h] [a] On the graph in Figure 6.67,
find five disjoint paths (no two paths have
any edge or vertex in common) so that the first
connects the two points labeled A to each
other, the second connects the points labeled
B, and so on.
D
E
1
1
C
B
B
C A
FIGURE 6.67
6.38. [s] [a] A builder is constructing three
houses on adjacent lots. He wishes to draw a
plan showing how to connect each house to the
gas company, the electric company, and the
water authority. (See Figure 6.68.)
Can he do this without any of the nine
lines crossing each other or passing through
any house or utility?

SOLVE IT WITH NETWORKS: AN INTRODUCTION TO GRAPH THEORY
211
1 Gas 1 1 Electric | | Water |
A
B
0
people will be in the same threesome on more
than one day.
Can you help them?
FIGURE 6.68
M 6.39. [s\ If there are six people at a
party, prove that at least three are mutual
acquaintances (they all know each other) or
else three are mutual non-acquaintances (none
of the three knows either of the other two).
([63], Problem 151)
6.40. [h] [a] Amos, Burl, Crawford, Dirk,
and Everett were seated, in that order, around
a circular table, playing poker. They took a
break to watch the fights on TV. When they
resumed playing, they changed their seats, so
that no one was now sitting next to a person
next to whom he had sat before.
What was the new seating arrangement?
6.41. [h] Seven women, Flora, Gail, Hala,
Iris, Jaclyn, Kitty, and Leona, were seated, in
that order, around a circular table, playing
poker. They took a break to watch the fights
on TV. Then they resumed playing. At
midnight, they interrupted their game again, for a
snack. When they resumed playing again, it
was observed that each woman had new
neighbors. In fact, each pair of women had
been neighbors during one of the three
segments of the card game.
Find seating arrangements that would
satisfy the stated conditions.
6.42. Monique, Nelson, Olivia, Patrick,
Queenie, Rosa, Sharim, Theo, and Una plan to
set out in teams of three to canvass a
neighborhood for a polling organization. They intend
to repeat this procedure for a total of four
consecutive days (counting the first). They
wish to arrange the teams so that no two
M 6.43. [h] [a] The eight queens on the
chessboard in Figure 6.69 dominate the board
in the sense that every cell not occupied by
a queen is under attack by at least one of
the queens. On the other hand, no two of the
queens attack each other. The problem is to
move exactly three of the queens so that, in
the final position, the queens will again
dominate the board and no two queens will
attack each other. (A queen may move any
distance horizontally, vertically, or
diagonally.)
Q
Q
Q
Q
Q
Q
Q
I IqI I
FIGURE 6.69
6.44. [h] It is possible to place five queens on
an 8 X 8 chessboard so that they dominate
the board (see Exercise 6.43). In the diagram
in Figure 6.70, four of these queens have

212
CHAPTER 6
already been placed. Place the fifth one, and
show that the answer is unique. ([48], pp.
206-207)
Q
Q I
Q
I IQ
FIGURE 6.70
6.45. [h] (a) Prove that on an w x w
chessboard it is not possible to place more than n
queens so that they are independent (that is, so
that no two queens attack each other).
(b) On the board in Figure 6.71, four queens
have already been placed. Place four more so
that the resulting eight queens are
independent. ([48], p. 211)
Q
Q
Q
1 Iq
FIGURE 6.71
6.46. [h] [a] What is the largest number of
knights that can be placed on a 6 x 6
chessboard so that they are independent? (See
Exercise 6.45.)

Games of Strategy
for Two Players
The word "game" is derived from the Old English word, gamen,
which means sport or amusement. In this sense the recreational
problems we have considered so far are games—games with one player,
you, the solver.
In this chapter we consider games involving two players. Some such
games were played prior to 2000 bce. Many old games have weathered
time or, in some cases, the rules have been altered. For example, the
game of chess, in one form, was known to the Hindus and Persians in
the sixth century. However, it did not come into its present form
until the sixteenth century.
On the other hand, new games are constantly being developed.
Probably the best sources of information in regard to recent games, as
well as old ones, are Martin Gardner's column, "Mathematical
Games," which appears in Scientific American, and his books ([18] to
[25]). In addition, several important new books on the subject have
recently appeared (for example, see [6] and [9]).
The determining factor in solving the problems we have encountered
in earlier chapters has been our ability to reason. In a two-person game,
another factor seems to be introduced—the reasoning ability of an
opponent. In some games, such as bridge, this is of great importance.
There is a class of games, however, for which the opponent's ability
may (at least theoretically) be neglected completely. The games that
we will investigate fall into this category. Some of these games will be
easy to analyze (that is, determine whether either player can force a
win), while others are mind boggling.
Many games of this type can now be played against a computer.
213

214
CHAPTER 7
Your analyses of these games will tell you whether or not you can
beat the computer, and, if so, what your strategy should be. Also, the
logic of the analyses can be quite useful in programming the computer
to play.
The following problems present a few games of the type we consider
in this chapter. See if you can analyze them on your own; then read
on for a discussion of some of the principles involved in conducting
an analysis. We suggest that if you find the analysis difficult at first, you
should play the game several times and, toward the end of each round,
start asking questions such as "What would have happened if I had
made a different move?"
SAMPLE
PROBLEMS
Problem 7.1
Two players. Thinker and Patzer, are about to play the following
game. From a given pile of matchsticks, each player, in turn, selects
any number of sticks up to some fixed maximum, thus gradually
depleting the pile. The player who selects the last stick is the winner.
If Thinker has the choice of whether or not to go first, tell which
he should choose and what strategy he should follow if
(a) there are 28 matchsticks in the pile and each player may select 1
or 2 sticks at each turn;
(b) there are 28 matchsticks in the pile and each player may select 1,
2, or 3 sticks at each turn;
(c) there are n matchsticks in the pile and each player may select
1, 2, ..., w — 1, or w sticks at each turn.
Problem 7.2
Consider the following game played on a 3 x 3 checkerboard. Two
players alternate turns placing checkers on the board. Each player, in
turn, may place checkers on as many of the vacant squares as he or
she wants to, provided that all checkers placed during a given turn
lie in the same horizontal row or the same vertical column. The
winner is the player who places the final (ninth) checker.
Is it better to go first or second? ([38], Chapter 1, Problem 39)
Problem 7.3
The game of Nim is played as follows: It begins with three (or more)
piles containing arbitrarily specified (usually unequal) numbers of
matchsticks. The players alternate turns. Each player, in turn, may
select as many sticks (up to the limit of the pile) as desired from

GAMES OF STRATEGY FOR TWO PLAYERS
215
any one pile in which there remain any sticks. He or she may not
select from more than one pile on any one turn. The player who
removes the final stick from the final pile is the winner.
(a) One commercial version of the game begins with piles of 3, 5,
and 7 sticks respectively. In this version of the game, is it best to go
first or second, and what strategy should you follow?
(b) Generalize to the game beginning with three piles containing />,
^, and r sticks respectively.
Problem 7.4
Given 15 dots on a line,
two players take turns placing an X through one of the dots. The first
player to mark off a dot so that three consecutive dots are marked
is the winner.
Which player should win—the one who goes first or the one who
goes second?
For each game that we consider in this chapter, there will be two
players, Alfie and Bette, whom we will generally refer to as A and B.
They play or "move" alternately, A moving first.
In most games, at his or her turn, a player may have to choose one
of several moves. A strategy is a rule or decisionmaking formula that
tells the player which choice to make at each turn. If this strategy
enables the player to win no matter what moves his or her opponent
makes, we will call it a winning strategy for that player. If the game
can end in a draw, we can also speak of a drawing strategy—a
strategy which, although it does not guarantee a win for the particular
player, guarantees that he or she does not lose.
For each game we ask the following question: "Does A have a
winning strategy?" If not, we ask whether B has one. If the answer
to either question is yes, we try to find this winning strategy. It is not
possible for both players to have winning strategies. And, of course, it
might be that neither has.
CHANCE-FREE DECISIONMAKING
In this chapter we ignore games in which "Lady Luck" plays a role
(poker, backgammon—games in general in which a player's fortune
depends on the roll of dice or the deal of cards). The analyses of games
such as these are far too complicated, involving lengthy probabilistic

216
CHAPTER 7
arguments, and, even if successfully completed, yield only strategies
by which one can optimize the chances of winning but cannot guarantee
a win. We will investigate only games in which each player at each
turn is free to choose any legal move—games in which decisions are
not made by chance.
GAMES OF PERFECT INFORMATION
As a second restriction, we are interested here only in games of
perfect information—games in which both players are aware at all
times of all aspects of the structure of the game. Each knows, at any
point in the game, what moves have been made prior to that point as
well as what moves the opponent will be able to make in response to
any possible move. Consider Scrabble for example. Aside from being
a game of chance (chance determines what letters you pick), it is also
a game of imperfect information, for you do not know what letters your
opponent holds. As a result, you might be reluctant to place certain
letters on the board for fear that doing so would open excellent scoring
opportunities for your opponent, opportunities of which he or she might
in fact not be able to take advantage. In general, not having full
information about a game introduces an element of chance. As a result,
it may not be possible for either player to develop a strategy that
guarantees a win—although it may be possible to find a strategy that
optimizes one's chances.
Consider, for example, the following game. Player A secretly selects
three numbers from 1 to 9. B then announces two numbers from 1 to 9.
If the sum of at least one of B's numbers with at least one of A's
numbers is 10, then B wins. Otherwise, A wins. (For instance, if A
picks 3, 5, and 7 and B picks 1 and 4, then B wins because
1 + 4 + 5 = 10; similarly, B wins if she chooses 1 and 2 because
2 + 3 + 5 = 10.) Although chance is not involved in the
decisionmaking process of this game (each player is free to choose whatever
number he or she likes), the lack of information about the opponent's
move makes luck a factor in the game. Neither player can adopt a
strategy that will guarantee a win.
FINITENESS
There is one other requirement we place on the games we consider
here—they must be finite. A finite game is one that must necessarily
terminate in a finite number of moves. Many board games, such as
backgammon and Monopoly, are not finite; they could conceivably
continue forever. As another example of an infinite game, consider the
following: A and B alternate turns at rolling a die. With each roll, the
player who rolled adds the resulting number to his or her previous

GAMES OF STRATEGY FOR TWO PLAYERS
217
total. The game ends when one player's total exceeds the other's total
by 20 points or more. This game could conceivably end after A's fourth
turn; or it could continue indefinitely. Frequently, we can make a game
finite by the imposition of a rule that limits play. For example, in the
game above, we could add a rule which says that if no player has
won by the end of 1000 rolls, then B is the winner.
A more interesting example of a rule that forces a game to be finite
occurs in chess. The rule says that if 50 consecutive moves are made
without a piece being captured or a pawn being moved, then the game
is a draw. Since each side can have at most a finite number
(8 • 6 = 48) of pawn moves and can make at most a finite number (15)
of captures, the game must terminate in a finite number of moves—at
most, 2(48 + 15) • 50 = 6300, since the game ends immediately if only
the two kings remain.
If there is a number, w, such that the game cannot last for more than
n moves, then we say that the game is bounded. If n is the smallest
such number, then we say that the game is of length n. Although
there are finite games that are not bounded (for example, player A
chooses a number and then A and B alternate turns rolling a die until
the total exceeds the chosen number), we do not consider such games
here. All bounded games have some fixed (maximal) length w, although
it may be difficult to say what that length is.
. [a] Consider each of the following games: Is it chance-free; is it a
game of perfect information; is it finite?
(a) There are 6 matchsticks on the table. Each player at a turn takes
1 or 2 matchsticks. The one who picks the last stick wins. To
see who goes first, the players flip a coin.
(b) The first player, A, writes any positive integer on a piece of paper,
then B writes one. If the sum of the two numbers is odd, A
wins. Otherwise, B wins.
(c) Backgammon.
(d) Gin rummy.
(e) Chinese checkers.
(f) Tic Tac Toe.
(g) Parcheesi.
PRACTICE
PROBLEMS
7.A
THE EXISTENCE OF WINNING STRATEGIES
If we now impose the above restrictions on a game, we have the
following result:

218 CHAPTER 7
Theorem 7.1 In any finite two-person game of perfect information
in which the players move alternately and in which chance does not
affect the decisionmaking process, either
(a) one of the two players must have a winning strategy
or
(b) the game is a theoretical draw (both players must have drawing
strategies).
To see this, observe that, since the game is finite, it must end in a win
for one player or in a draw. If A does not have a winning strategy,
then, whatever play A makes, B must have a counterplay to prevent A
from winning (that is, if there is a play that A can make which cannot
be countered by B, then that play would be part of a winning strategy
for A); thus, B has at least a drawing strategy. It may be that B
actually has a winning strategy. If not, then it must be possible for A
to play in such a way that, no matter what B does, B does not win;
in other words, A has a drawing strategy.
Thus, one of the players has a winning strategy or both players have
drawing strategies.
Note that, if the game cannot end in a draw, then one of the two
players must have a winning strategy—preventing your opponent from
winning is equivalent to winning yourself.
A formal proof of Theorem 7.1 for games of finite length can be
given, using induction on the length of the game. We do not present the
complete details of the proof, but we illustrate the inductive step by
the following example. (For a discussion of mathematical induction,
see Appendix B.)
Consider a game in which A selects a number, either 3 or 4; B
adds either 3 or 4 to A's number; A then adds 3 or 4 to the
total; etc. If either player brings the total to 13, he or she wins; if the
player brings the total to 14, the game is a draw; if the player brings
the total over 14, that player loses.
The tree diagram of the game is shown in Figure 7.1. Since the
game ends in at most five moves, the game is of length 5.
Consider the nodes from which all moves end the game immediately
(in Figure 7.1, these nodes are circled). If such a node is reached in
the course of play, it should be clear who will win. Specifically, if it is
A's turn at such a node and if one of A's possible choices results
in a win for A, then obviously A will make that choice and will win.
If none of A's choices result in a win for A but one of them results
in a draw, then A will make that choice and the game will be drawn.
If A cannot make a choice that results in a win or a draw, then B
will win if that node is reached. The same principles apply if B is
to make the choice. Since the outcome of the game is strategically
determined when a node of the type in question is reached, we can

GAMES OF STRATEGY FOR TWO PLAYERS
219
STRATEGIC
A, B, A2 B2 A3 RESULT
3_^-15...Bwins
4"~~"~^16...B wins
/ 9 r
4^""~^13 B wins
.6
-. / ^\^ ^^-"^ 13 B wins
4"~~~~"14 draw
^3
3_,_,—-13 B wins
4\ ^'°~ ^
3/ \ ^^^^ 4^~~~~^14 draw
7.
3 --14 draw
4~"~~^15 A wins
0
3,^^-13 B wins
4^""'~^14 draw
4\ ->/ ^^^ 3^^---14 draw
4~~ 15 A wins
3_.--'14 draw
*+ \ ^ 4^^^15 Awins
3,,,--—15 A wins
4~~"~^16 A wins
A, represents A's ith move; B, represents B's ith move; the number at the node
indicates the total at that point; the number above each line indicates the number
selected at that turn.
FIGURE 7.1
replace such nodes by the associared srraregic results. Doing this for
the tree in Figure 7.1, we obtain Figure 7.2.
This figure represents a game of length 4. In general, the length will

220
CHAPTER 7
STRATEGIC
A, B, A2 B2 RESULT
3^.^,.^ 12 .8 wins
^^ 4^~^^ 13 .. B wins
^y^ 4^^^ 10 Bwins
/3^
^\^ 3....^-10 Bwins
4"~~^ 11 draw
0
\3^^^ 10 B wins
^^ 4"^"^^-^ 11 draw
4
^\^ 3,-^- 11 draw
4"""~~^ 12 A wins
FIGURE 7.2
be decreased by 1 and we could apply the induction hypothesis; that
is, if we assume that a suitable game of length n satisfies the conclusion
of Theorem 7.1, we would know that each such game of length n -I- 1
is reducible to a game of length w, and therefore also satisfies the
conclusion of the theorem. The circles in Figure 7.2 indicate where
nodes were eliminated from the previous diagram.
Although it would not be necessary to do so in the inductive proof,
let us continue the process for our example: We consider next the
nodes indicated by boxes in Figure 7.2. Replacing them by their
strategic equivalents, we obtain Figure 7.3a.
A „ ^ STRATEGIC
^1 B, A2 RESULT
^^X---^ ^ • B wins
^^ 4"^ 10...B wins
^y^ 4 " draw
^\
^\. ^^^^-^""^ ^ draw
4 8 A wins
(a)
FIGURE 7.3

GAMES OF STRATEGY FOR TWO PLAYERS 221
Continuing this process further, we obtain Figure 1.3b,
S^,^^" 6 B wins
3
0 ^
4
4^ 7 draw
4 draw
(b)
then Figure 7.3c,
(c)
and finally Figure 7.3d.
0 draw
(d)
Thus, the game is a theoretical draw.
In the analysis of the game, we have associated a value with each
node. Returning to the original tree diagrarn, we fill in these values,
obtaining Figure 7.4 (next page).
From this diagram, each player's strategy should be clear. A must
always play to a node whose strategic value is "A wins" or "draw,"
and B must always play to a node whose value is " B wins " or " draw."
If a player deviates from this strategy, he or she may lose.
A word of caution: In general, drawing a tree diagram is not always
a reasonable approach to finding strategies—the diagrams are usually
too immense. We will return later to the question of how to go about
trying to find a winning strategy.
Theorem 7.1 says that in every game of the type under consideration,
either one player has a winning strategy or both players have drawing
strategies. But determining which of these cases applies may not be
easy. For some games, such as checkers and chess, the determination
has never been made. In these games. Theorem 7.1 implies that either
the first player has a theoretical (forced) win, or the second player has
a theoretical win, or the game is a theoretical draw; but the vast number
of possible sequences of moves in these games has enabled them to
withstand complete analysis.

222
CHAPTER 7
A.
U-draw
J ^15-B wins
'12-B wins
4^^^ 16-B wins
3-Bw"
9 Bwins
4^^13-B wins
J 13-Bwins
10^ wins
4 14 dra
^ --13 Bwins
Kf^ wins
4 """^14 draw
^ .--- 14 -draw
11 rdraw
4^^"'^1'^ A wins
J ^ 13 B wins
10^ wins
4 14 draw
14 drav
FIGURE 7.4
Even if it is possible to tell which player has the theoretical win,
it may not be easy to find the winning strategy. That is, it may be
possible to prove that a winning strategy exists without actually being
able to find such a strategy. As an example, consider the game of Hex
(see [18], p. 73), which was invented in the 1940s by the Danish
engineer, Piet Hein. The game is played on a diamond shaped board
made up of hexagons: A, playing white, and B, playing black, take

GAMES OF STRATEGY FOR TWO PLAYERS
223
turns placing markers, one at each turn, on vacant hexagons. A's object
is to complete a connected path of white markers that touches the top
and the bottom of the board. B's objective is similar, except that her
path of black markers must go from side to side. Whoever completes
such a path first is the winner.
Figure 7.5 illustrates a completed game of Hex on a 6 x 6 board.
B (black) has won; her winning path is indicated by the arrows.
White
Black
Black
FIGURE 7.5 White
It is fairly easy to show that the first player has a winning strategy
for any size Hex board (n units on a side). To see this, we must first
see that the game cannot be drawn. Suppose a draw were possible,
yielding a board that is completely filled, but no one has won. Imagine
taking a pair of scissors and cutting out all cells on which a black
marker is found. Since A has not won, the board must separate into
pieces so that no one piece contains cells from both the bottom and
top of the board (otherwise the piece would contain a winning path
for A). Therefore, there must be a series of cuts that completely

224
CHAPTER 7
separates the top of the board from the bottom. These cuts must go
from the left side of the board to the right side, and hence contain
a winning path for B.
Since the game cannot end in a draw, either A or B must have a
winning strategy. If B had the winning strategy, then A could make any
move he likes and then, ignoring his first move, play as if he were
the second player. The extra marker he has on the board can in no way
hinder him; nor can it help B. And if at some stage in the game the
strategy calls for A to place a marker in the cell he originally marked,
since he already has a marker there, he can again make any move he
likes. Thus, A will win—yielding a contradiction to the assumption
that B has a winning strategy. Hence, A, the first player, must be the
one who has the winning strategy.
This same argument works for any game that cannot be drawn and
in which a player's earlier moves do not limit his or her possible later
moves and cannot help the opponent.
Even though it is known that the first player in a game of Hex has a
winning strategy, it is still not known, for large boards, what that
strategy is.
PRACTICE
PROBLEMS
7.B
1. [a] In the game of Tic Tac Toe, a player's first move cannot hurt
him or her later on. Why, then, doesn't the argument used above
for the game of Hex show that A has a winning strategy for Tic
Tac Toe?
2. [a] Show that B cannot have a winning strategy in the game Qubic
(a three-dimensional, 4x4x4 version of Tic Tac Toe) where the
object for A is to get four X's in a row and the object for B is
to get four O's in a row.)
3. Consider the following game played on a 4 x 4 x 4 Tic Tac Toe
board. Each player at his or her turn places an X on any cell of the
board. The first player to complete four X's in a row is the winner.
Does an argument similar to the one used for Hex—to show that
the first player has a winning strategy—apply to this game? Explain.
4. In the game of Bridg-it (see page 262), prove that the first player
must have a winning strategy.
POSITION-STATE OF THE GAME
Before we begin to analyze games, that is, determine if winning
strategies exist, it is convenient to be able to refer to the "state of the

GAMES OF STRATEGY FOR TWO PLAYERS 225
game" or "position" at any point during play. When we do so, we
may include all of the following factors:
1. The physical placement of the pieces on the board (if it is a board
game) or some analogous notion (if it is not a board game).
2. Which player is to make the next move.
3. What moves are legal for that player to make at the time in question.
4. What moves have been made up to that time.
We may think of a strategy as a chart listing all possible positions
that could arise during the play of the game together with a description
of what action should be taken (what move should be made) in each
position.
The position that exists just prior to the start of the game will be
referred to as the initial or starting position of the game. A position
in which the game is over will be called a terminal position. Each
nonterminal position that could conceivably arise in the course of play
may be regarded as the starting position of a different, but related,
game. Hence, by Theorem 7.1, from each particular possible position
in the game, either the player who is about to move can force a win,
or the opponent can force a win, or both players can force a draw.
We adopt the viewpoint of the player, X, who has just moved,
thereby "leaving the game in a particular position." If that position is
one from which X can force a win, then the position is obviously
a favorable or winning position (for X). Similarly, if X has left a
position from which his or her opponent can force a win, then that
position is an unfavorable or losing position. A position from which
both players can force a draw is called a drawing position. We can
thus classify each position that could arise in a game as a winning,
a losing, or a drawing position—for the player who has just moved.
(Obviously, there can be no drawing positions in games that cannot
end in a draw.)
A terminal position in which the player who has just moved has won
the game is clearly a winning position. Similarly, a terminal position in
which the game is drawn is a drawing position, and a terminal position
in which the player who has just moved has lost is a losing position.
If X leaves a nonterminal winning position (from which he or she
can force a win), then X's opponent, Y, on the next turn, will be
unable to leave a position from which Y too could force a win, or even
a draw; otherwise the statement that X could force a win would not
be true. In other words, Y's move will have to result in a losing position
(for Y). Thus, any move made from a winning position must result in
a losing position.
Similarly, if X leaves a nonterminal losing position, then it must

226 CHAPTER 7
be possible for Y to force a win by choosing moves properly. In other
words, from any losing position, there must be at least one move
possible that results in a winning position.
Finally, if X leaves a nonterminal drawing position, then it must be
impossible for Y to force a win but it must be possible for Y to force
a draw. Thus, any move made from a drawing position must either
leave a drawing position or a losing position, and at least one possible
move will actually leave a drawing position.
In summary: The possible positions in a game in which draws are
possible may be divided into three lists—I, winning positions; II, losing
positions; III, drawing positions—such that the following conditions
are satisfied:
1. Terminal positions, if any, in which the player who just moved has
won are on list I; terminal positions, if any, in which the player
who just moved has lost are on list II; terminal positions, if any,
in which the game is drawn are on list III.
2. Any move made from a position on list I results in a position on
list II.
3. From any nonterminal position on list II, there is at least one move
possible that results in a position on list I.
4. Any move made from a position on list III results in a position in
list II or III.
5. From any nonterminal position in list III, there is at least one move
possible that results in another position in list III.
If draws are not possible, we get only two lists that satisfy conditions
1, 2, and 3.
Once we know the winning, losing, and drawing positions for a
game, the game has been analyzed and it is easy to tell which player
(if either) can force a win and to describe the winning or drawing
strategy.
If the starting position is a losing position, then A can win as
follows: His first move must be chosen so that he leaves a winning
position. This is possible by condition 3 above. Thereafter, every
move B makes will leave a losing position (condition 2), and A will
always be able to choose his move so that he leaves a winning position
(condition 3). Since we are dealing with a finite game, a terminal
position must eventually be reached. Since B always leaves losing
positions and A always leaves winning positions, A will win regardless
of who reaches the terminal position: If B reaches it, it must be a
terminal position in which the player who just moved lost; whereas, if
A reaches it, it must be one in which the player who just moved won.

GAMES OF STRATEGY FOR TWO PLAYERS
227
Similarly, if the starting position is a winning position, then B
has a winning strategy. Each time A moves, he will leave a losing
position (condition 2), and each time B moves she will be able to leave
a winning position (condition 3).
Finally, if the game starts at a drawing position, then both players
have drawing strategies. A must play so that he leaves a drawing
position (this is possible by condition 5); then, so must B; and so on.
Any deviation from this strategy will result in a player leaving a
losing position (condition 4), from which it will be possible for his
or her opponent to win.
Thus, we can analyze a game by determining whether each possible
position that could arise is a winning, losing, or drawing position.
Actually, though, it is not always necessary to investigate all possible
positions. As long as we can find three lists of positions that satisfy
the five conditions above (or two lists that satisfy conditions 1, 2, and 3)
such that the starting position is on one of the lists, the same reasoning
used above enables us to determine which player wins and what
strategy he or she should follow. We illustrate with the following
example.
An Example—Another Matchstick Game
From a pile of 11 matchsticks, two players take turns removing either
1 or 4 sticks. The player who removes the last stick is the winner.
In this game, a position may be represented by a number—the number
of matchsticks remaining. (Note that no draws are possible in this
game and that the only terminal position, 0, is a winning position.)
Ignoring for the moment the question of how the lists in Figure 7.6
were obtained, it is easily verified that they do satisfy conditions 1,
2, and 3 above, and hence they give the winning and losing positions
for the game.
From any position in the winning column, subtracting 1 or 4 gives a
position in the losing column. Similarly, from each position in the
losing column, it is possible to obtain a position in the winning
column either by subtracting 1 or by subtracting 4, as the case may be.
Since the starting position, 11, is a losing position, A wins this game
(if he plays properly). He may select either 1 or 4 sticks and thus
place himself in a winning position. Whatever B then does will leave
a losing position, from which A will be able to return to a winning
position, etc. To fully describe A's strategy, we must say what he
should do in each possible case. This is obvious from the chart.
The chart in Figure 7.6 contains all possible positions that could
arise in the game. Actually, however, shorter lists of winning and
losing positions are possible. For example, the lists in Figure 7.7 also
satisfy conditions 1, 2, and 3; and A's strategy using these Hsts is
winning
10
7
5
2
0
losing
11
9
8
6
4
3
1
FIGURE 7.6
winning
7
2
0
losing
11
6
3
1
FIGURE 7.7

228
CHAPTER 7
easily stated: On his first turn, he should leave 7 sticks; on his second
turn, he should leave 2 sticks (this will be possible no matter what
B's first move is); on his third turn, he wins. The reason that we
were able to omit some possible positions from the chart is that A
was able to adopt a winning strategy in which the omitted positions
could not occur.
Note that the choice of which winning positions we include on our
list must relate to which losing positions are included, and vice versa;
conditions 1, 2, and 3 must be satisfied. Hence, if we include 10
in our list of winning positions, then 9 must be included in the list of
losing positions, so that 5 would have to be on the list of winning
positions (8 cannot be since it is a losing position), and so on.
It is important to observe that although the game in question is a
winning game for A, he can lose if he plays poorly. If he begins by
taking 1 stick and then again takes 1 stick after B takes 1 stick, then he
leaves 8 sticks—a losing position, and B can win with proper play.
As a final comment in this section, we note that it would not be
correct for A to reason as follows: ''Since 2 is a winning position, I
can make random moves until there are a little more than 2 sticks left,
and then I will be sure to leave 2 sticks."
B may be able to prevent him from leaving 2 sticks. In other
words, in a game of the type we are considering in this chapter, a
strategy must start from the very first move; it cannot begin in the
middle of the game. In the game at hand, A must plan his strategy
so that he is sure of being able to leave 2 sticks.
PRACTICE
PROBLEMS
7.C
Analyze the matchstick game above by making a tree diagram and
labeling the nodes as in Figure 7.4. Who will win? Find the
winning strategy. Compare your answer with the charts in Figures
7.6 and 7.7.
THE STATE DIAGRAM OF A GAME
In addition to the tree diagram, there is another graph-theoretical
approach to analyzing games. It is the state diagram of a game. In
this diagram, each node represents a position or state of the game.
A directed edge is drawn from node P to node Q if it is possible
to move from position P to position Q in one move. For example,
the state diagram of the game discussed in the preceding section is
shown in Figure 7.8. Note that from the starting position, 11, we can
go to either 10 or 7; from 10, we can go to 9 or 6; etc.

GAMES OF STRATEGY FOR TWO PLAYERS
229
FIGURE 7.8
In the approach, sometimes referred to as the Sprague-Grundy
method, the nodes of the diagram are labeled as winning, losing, or
drawing, by starting at the terminal nodes and working backward.
In this example, 0 is a winning position, so we label it W. We
now work backward, using our definition of winning and losing
positions. (This game cannot end in a draw, so drawing positions need
not be considered.) Since 0 can be reached by a move from 4 or from
1, these last two positions must be losing; so we label them L
(Figure 7.9).
We continue, following the definitions of winning and losing
positions: That is, a node from which at least one edge leads to a
position labeled W must be labeled L; if all edges from a node lead
to other nodes labeled L, then the node in question is labeled W.
(In a game that can end in a draw, if some edge from a node leads
to another node labeled D and no edges lead to nodes labeled W,
then the node in question should be labeled D.)
For the game in question, we obtain Figure 7.10.
►0 w
FIGURE 7.9
FIGURE 7.10
1. Use the Sprague-Grundy method to analyze the game for which the
tree diagram is shown in Figure 7.1 (page 219).
2. Use the Sprague-Grundy method to analyze the matchstick game
in Sample Problem 1.5 (page 3).
PRACTICE
PROBLEMS
7.D
Z^_D

230
CHAPTER 7
HOW DO WE FIND A WINNING STRATEGY?
Theoretically, the tree diagram approach used to analyze the game
earlier in the chapter could be used to find a winning (or drawing)
strategy for any game of the type we consider. For many games, the
state diagram could also be used. However, with most games, there are
just too many possible positions to make it practicable to draw a tree
or state diagram. We therefore now investigate some other approaches
that are often helpful in analyzing a game as well as some techniques
for limiting the number of cases that need be considered.
Solution of
Problem 7.1
FINDING A WINNING STRATEGY BY WORKING
BACKWARD
In the tree diagram approach illustrated earlier, we were able, by
working backward, to assign a value (A wins, B wins, or draw) to
each node or position that could arise in the game. Similarly, in the
state diagram approach, we again worked backward to assign a value
W, L, or D to each position. Using this information, we were then
able to find each player's proper strategy.
In general, even if we have not drawn a tree or state diagram of
the game, we may still be able to work backward to determine that
certain positions are winning or losing positions.
For example, we can analyze the game in Problem 7.1 by working
backward, without drawing a tree or state diagram of the game.
If you have not already solved Problem 7.1 (page 214),
try it again now.
If, in playing this game, 0 sticks remain after a player's turn, that
player has won; so 0 is a winning position. (Note that, in this game,
a position may be represented by the number of sticks remaining.)
If 1 or 2 sticks are left after a player's turn, that player will lose,
as the other player may take the remaining stick(s). Thus, 1 and 2
are losing positions.
Continuing in this manner, we obtain Figure 7.11 for part (a) of the
problem, in which each player may remove only 1 or 2 sticks at a turn.
inning
0
3
6
losing
1,2
4,5
7, 8
reason
goal has been achieved
opponent can take last siick(s)
opponent must leave 1 or 2 (losing)
opponent may leave 3 (winning'
opponent must leave 4 or 5 (losing^
opponent may leave 6 (wmnmg^
FIGURE 7.11

GAMES OF STRATEGY FOR TWO PLAYERS
231
We can also represent this as in Figure 7.12.
number
remaining
0
1
2
3
4
5
6
7
8
winning
or losing
winning
losing
losing
winning
losing
losing
winning
losing
losing
FIGURE 7.12
If necessary, we could continue all the way up to 28; but you should
notice a pattern developing. The numbers that are divisible by 3 seem to
be winning positions, and all other numbers seem to be losing
positions. To verify that this is in fact the case, we must show that
conditions 1, 2, and 3 for a game (page 226) are satisfied by the
two lists:
List I: All numbers divisible by 3 (of the form 3k).
List II: All numbers not divisible by 3 (of the form ?>k + \ or
3k + 2, depending on the remainder upon division by 3).
Note that the only terminal position in this game, 0, is a winning
position.
Since 0 = 3 • 0 is divisible by 3, condition 1 is satisfied.
If 1 or 2 is subtracted from a number divisible by 3, the result
will not be divisible by 3 [3/j - 1 = 3(/j - 1) -f 2 and 3k - 2 =
3(k — 1) -f 1]. Hence, condition 2 is satisfied.
Taking 1 from 3/j -f 1 or 2 from 3/j -f 2 leaves 3k, a number
divisible by 3; hence, condition 3 is satisfied.
Thus, in Problem 7.1(a), Thinker's strategy is clear. He should go
first, taking 1, leaving 27 sticks. Thereafter, he should always leave a
multiple of 3.
Note that our discussion above has more than answered Problem
7.1(a). It has also solved Problem 7.1(c) in the case that m = 2.
Problem 7.1(b) may be solved similarly. Working backward, we
obtain the lists of winning and losing positions shown in Figure 7.13.
Again we are led to a conjecture: The winning positions are the
multiples of 4 (numbers of the form 4k); all other positions (numbers
of the form 4k -\- 1, 4/j -f 2, or 4/j -f 3) are losing positions. [Can you
prove this? Try it using the method used in part (a).]
Thus, in Problem 7.1(b), Thinker should choose to go second, and
should always leave a multiple of 4 sticks.
Can you now generalize the method to find a solution of Problem
7.1(c)? Problem 7.1(a) and 7.1(b) should suggest a conjecture for the
FIGURE 7.13

232
CHAPTER 7
winning positions in the general game. Or, if not, you should be able
to make one by examining the lists of winning and losing positions in
Figure 7.14.
winning
0
m + 1
2(m + 1)
m
losing
1,2, . . . ,m
^ 2, . . . ,2m + 1
reason
goal has been achieved
opponent may take last stick
opponent must leave 1,2,...
opponent may leave m + 1
opponent must leave m f 2, .
, or m
. . , or 2m + 1
FIGURE 7.14
Conjecture For the game in Problem 7.1(c), the winning positions
are all multiples of w + 1 [numbers of the form (m + 1)^] and the
losing positions are all other numbers [numbers of the form
(w + l)/j + r, with 1 < r < w, where r is the remainder when the
number is divided by w + 1].
We prove the conjecture by verifying conditions 1, 2, and 3 (note
that, again, 0 is the only terminal position):
1.0 = (w + 1) 0
2. If we take t sticks, \ < t < m, from a multiple of w + 1, we do not
get a multiple of w + 1. That is, for I < t < m,
{m+ \)k- t = {m+ \){k - 1) + (w + 1 - 0,
where \ <m + \ — t <m.
3. We can take r sticks from (m + \)k + r to leave (m + \)k.
Thus, Thinker should always leave multiples of m + 1 sticks.
Whether he should choose to go first or second depends on n. If n is
a multiple of w H- 1, Thinker should go second; otherwise, he should
go first.
PRACTICE
PROBLEMS
(5^12 7.E
1. Verify the conjecture for Sample Problem 7.1(b).
FINDING WINNING STRATEGIES BY
SIMPLIFYING A GAME
In the above solution of Problem 7.1, although we essentially used the
technique of working backward, we also used another technique—that

GAMES OF STRATEGY FOR TWO PLAYERS 233
of simplifying the problem. In working backward, we replaced the
original problem by a sequence of simpler problems:
A game with 0 sticks;
A game with 1 stick;
A game with 2 sticks;
Etc.
Each of these was easily solved from the previous ones.
In addition, we conjectured a solution of the general problem
[Sample Problem 7.1(c)] by first solving the special cases in 7.1(a)
and 7.1(b). In fact, if part (c) of this problem were presented without
parts (a) and (b), a good approach would be to consider special cases
such as (a) and (b), and then to generalize in order to find a solution
of part (c).
This technique of simplifying a problem was already encountered
in Chapter 1. In general, we can try to simplify a game by reducing
its size in one way or another.
For example, in Chapter 1, we solved the matchstick game that
started with 6 or 7 sticks. We found that the second player would win
in the 6 stick version and the first player would win when the game
began with 7 sticks. Thus, in the 28 stick game—or any version of
the game starting with at least 7 sticks—6 is a winning position and
7 is a losing position. Since at most 2 sticks may be taken at a time,
any play of the 28 stick game must reach a point at which either
6 or 7 sticks remain (it is impossible to bypass both of these
positions). When this point is reached, we need not continue playing,
for we know who will win. We have thus shortened Problem 7.1(a)
to the following: Starting with 28 sticks, play to leave exactly 6
sticks (and not 7). A little thought shows that this is essentially the
same as the problem," Starting with 22 sticks, play to take the last one."
This in turn can be reduced to, "Starting with 16 sticks, play to take
the last one." And so on until only the 4 stick game remains to be
analyzed.
Thus, we can solve the 28 stick game by being able to solve the
6 stick game and the 4 stick game.
FINDING WINNING STRATEGIES WITH A
FRONTAL ASSAULT
Although the procedure of working backward to reduce a game is an
extremely important one, there are times when it is not much help.
For example, if there are a large number of possible penultimate
positions (positions from which the game can end in one move), then
there is little to be gained by working backward rather than making
a frontal assault.
We should therefore consider all possible opening moves that the
first player could make, all possible responses by the second player.

234
CHAPTER 7
and so on. The Multiplication Principle makes it clear that the
number of possible cases grows very rapidly as the game progresses.
For example, if A has five possible choices for his first move and
B may respond to each of these in any of four ways, then we already
have twenty possibilities to examine. If A may make his next move in
either of two ways, then there are forty branches in a tree diagram.
And so on. Fortunately, there are several factors that may limit the
number of possible cases which must be investigated.
HOW MANY POSSIBILITIES NEED BE
CONSIDERED?
The number of possibilities we need investigate is limited by our
objective—to find a winning (or drawing) strategy. If we find that a
particular move by a particular player leads to a forced win for that
player, then we need no longer consider his or her alternative
moves. (We may wish to do so out of curiosity, but it is not essential.)
On the other hand, we cannot conclude that the player will lose until
we have investigated every possible move he or she could make.
SYMMETRY AS A LIMITING FACTOR
Another factor that can help limit the number of possibilities which
need be considered is symmetry.
Imagine the following situation: Mel and Ira are playing a game of
Tic Tac Toe (also known as Tit Tat Toe or Noughts and Crosses)
at a blackboard. Their younger brothers, Lem and Ari, cannot see the
blackboard directly but can follow the progress of the game through
a mirror placed on a neighboring wall. Lem and Ari decide that they
too will play Tic Tac Toe. In fact, Lem decides that he will always
make the move that Mel appears to be making and Ari decides that
he will always make the move that Ira appears to be making.
If the squares of Mel and Ira's
board were numbered as in
Figure 7.15a,
1
4
7
2
5
8
3
6
9
(a)
FIGURE 7.15
Thus, if Mel's first move is as in
Figure 7.16a,
then it would appear to Lem and
Ari that they were numbered as
in Figure 7.15b.
3
6
9
2
5
8
1
4
7
(b)
then Lem's first move will be as
in Figure 7.16b.

GAMES OF STRATEGY FOR TWO PLAYERS
235
(a)
FIGURE 7.16
(b)
It is clear that Lem will win his game if and only if Mel wins his,
and that Ari will win if and only if Ira does. Furthermore, information
about the appearance of either board during the game conveys
information about the appearance of the other board. For example,
if Mel and Ira's board appears
as in Figure 7.17a,
X
0
0
X
X
u
(a)
then Lem and Ari's board must
appear as in Figure 7.17b.
X
C)
0
0
X
X
(b)
FIGURE 7.17
Specifically, Lem and Ari's board may be obtained by flipping Mel
and Ira's board about a vertical axis.
Clearly, if at some time during the game Mel can force a win, then,
at that same time, Lem must also be able to force a win. A similar
statement holds for Ira and Ari's winning positions. Thus, the strategic
value of any position of Mel and Ira's board is the same as the
strategic value of the corresponding position on Lem and Ari's board.
In other words, if in analyzing the game we discover that a particular
position is a winning (losing, drawing) position, then the corresponding
reflected position is also a winning (losing, drawing respectively)
position, and we need not analyze it further. We simply say that the
two positions are equivalent "by symmetry."
Sometimes symmetries are easy to see. For example, in Tic Tac
Toe, we could show that the game has seven symmetries. Figure
7.18 shows one position and its seven equivalent positions.
We say that the positions in parts a and b of the figure are equivalent
by vertical symmetry (turned on the vertical axis); those in parts a
and c are equivalent by horizontal symmetry; etc. Symmetry
obtained by using 180° rotation, as in part g of Figure 7.18, is often
referred to as central symmetry.
If we determine the strategic value of any of the eight positions in
Figure 7.18, then the strategic value of each of the others is also known.
It follows that, in analyzing a game such as Tic Tac Toe, we do not
have to consider every possible sequence of moves. Symmetry enables

236
CHAPTER 7
0 X
(a)
vertical
flip
(b)
horizontal
flip
(c)
\ 1
\ X
0
X
\
\
\
\
\
1 \
diagonal
fliPi
(d)
/
diagonal
fliP2
(e)
[CURE 7.18
90°
rotation
(f)
180°
rotation
(g)
270°
rotation
(h)
us to limit the number of positions that must be considered in
analyzing the game. For example, as far as the opening move of the
game is concerned, we have to consider only the three possibilities
shown in Figure 7.19. Each other possible opening is equivalent to one
of these three by symmetry.
FIGURE 7.19
PRACTICE 1. [a] In Tic Tac Toe, which of the positions in Figure 7.20 are
pn^^PI pK^Q equivalent by symmetry?
7.F
(a)
(b)
FIGURE 7.20
(d)
(c)

GAMES OF STRATEGY FOR TWO PLAYERS
237
2. In Tic Tac Toe, which of the positions in Figure 7.21 are equivalent
by symmetry?
0
X
0
X
X
X
0
0
X
0
X
0
X
X
0
0
X
X
X
X
0
X
X
X
0
0
0
X
X
0
0
X
X
0
X
0
(a)
FIGURE 7.21
(b)
(c)
(d)
3. [a] Given the equivalent positions shown in Figure 7.22, what
move in part b of this figure corresponds to the move of placing an
X in the upper lefthand corner in part a?
X
X
0
X
0
0
(a)
FIG
UR]
E 7.22
0
0
X
0
X
X
(b)
4. Given the equivalent positions shown in Figure 7.23, what move in
part b corresponds to the move of placing an X in the upper lefthand
corner in part a?
X
0
X
X
0
0
X
X
0
(e)
(a)
FIGURE 7.23
Although symmetry is an intuitively easy concept in Tic Tac Toe,
in some games the various symmetries are not so obvious. In order to
elaborate on this point, we should be a little more precise in explaining
symmetry. Unfortunately, it is very difficult to give a precise definition
without using concepts that go beyond the scope of this book.
Therefore, we will rely on your intuition and simply say that two positions
are equivalent by symmetry if to each move from one of the positions
there is a corresponding move from the other so that the game
proceeds in essentially the same way from both positions. If there is a

238
CHAPTER 7
sequence of moves by which a particular player can force a win from
one position, then, from any equivalent positions, there will be a
corresponding sequence of moves by which that player can force a
win. In other words, two equivalent positions have the same strategic
values—they are both winning positions, both losing positions, or both
drawing positions.
Symmetry Applied to The Game of Problem 7.2
Whether or not two positions are equivalent depends basically on the
rules of the game. For example, consider the game in Sample
Problem 7.2 (page 214). This game is a variation of Tic Tac Toe. It
is not hard to see that the seven symmetries of ordinary Tic Tac Toe
are also present in this game. However, the nature of this new game
gives it additional symmetries that are not symmetries of Tic Tac Toe.
Observe that the only consideration governing the legality of a move
in the game of Problem 7.2 is whether or not certain squares are in the
same row or the same column. Clearly then, changing the order of
the rows or of the columns does not significantly alter the game. For
example, cutting the righthand column off the board and pasting it
onto the left of the board rearranges the board as indicated in Figure
7.24. For any move on the first board, there is a corresponding move
on the second board, which leaves the resulting positions equivalent.
Thus, for example, the positions in Figure 7.25 are essentially the
same.
1
4
7
2
5
8
3
6
9
3
6
9
1
4
7
2
5
8
• •
• •
FIGURE 7.24
(a)
FIGURE 7.25
(b)
The sequence of moves shown in Figure 7.26a corresponds to the
sequence in Figure 7.26b.
• •
• • •
• • •
• • •
• • •
• • •
• • •
(a)
FIGURE 7.26

GAMES OF STRATEGY FOR TWO PLAYERS
239
• •
• • •
• • •
• • •
• • •
• • •
• • •
FIGURE 7.26
(b)
Thus, reordering the rows or columns of the board gives a symmetry
of the game in Problem 7.2.
Does this also give a symmetry of ordinary Tic Tac Toe? No!
For example, the position shown in Figure 7.27a (in which X wins)
is clearly not equivalent to the position in Figure 7.27b (in which
O wins), which was obtained by moving the rightmost column to the
left.
0
X
0
X
X
0
X 0
I 0 I X
0 I X I
(a)
FIGURE 7.27
(b)
The reason that this type of symmetry does not apply to ordinary
Tic Tac Toe is that reordering the rows (or columns) changes the
diagonals, and the diagonals are important in Tic Tac Toe. On the
other hand, they are not important in the game of Problem 7.2.
We can conclude that different games may have different symmetries
even if they are played on the same board. In fact, even games that
are not board games may have symmetries (for example, see the
analysis of the game of Nim—Problem 7.3—below or the game of
Sprouts in the exercises, page 265).
In general, the greater the number of symmetries that may be
applied to a game, the more we can limit the number of positions we
need consider in analyzing the game. For example, although there are
many—thirty-three—possible opening moves that the first player could
make in the game of Problem 7.2, owing to the large number of
symmetries of this game, only three of them are actually different (see
Figure 7.28). All legal opening moves placing three checkers are
• •• •• •
(a)
FIGURE 7.28
(b)
(c)

240
CHAPTER 7
equivalent; all legal opening moves placing two checkers are equivalent;
and all opening moves placing one checker are equivalent. For example,
the position in Figure 7.29a may be seen to be equivalent to the
position in Figure 7.29d.
•
•
bottom
row to
top
•
•
right
column
to left
•
•
diagonal
flip
/
/
/
•
/
/
•
(a)
FIGURE 7.29
(b)
(c)
(d)
PRACTICE 1. [a] In the game of Problem 7.2, which of the pairs of positions
PROBLEMS Figure 7.30 are equivalent?
7.G
•
•
•
•
and
•
•
•
•
•
•
•
and
•
•
•
(a)
(b)
•
•
•
and
•
•
•
(c)
FIGURE 7.30
2. In the game of Problem 7.2, which of the pairs of positions in
Figure 7.31 are equivalent?
•
•
•
and
•
•
•
•
•
•
•
•
and
•
•
•
•
•
•
•
•
•
•
•
and
•
•
•
•
•
•
(a)
(b)
(c)
FIGURE 7.31

GAMES OF STRATEGY FOR TWO PLAYERS
241
DEjA VU-WE'VE SEEN IT BEFORE
For each of A's possible moves, B may choose a number of possible
responses. This brings us to a third factor that limits the possibilities
we need consider—deja vu (this is pronounced day-zha-voo, and is
French for "already seen")- If ^^he same position results from two
different sequences of moves, it need be considered only once when
analyzing a game. Thus, for example, if we find that the case in which
A moves as in Figure 7.32a and B responds as in Figure 7.32b leads
to a win for B, then the case in which A moves as in Figure 7.32c
and B responds as in Figure 7.32d also leads to a win for B. In fact,
this is true not only for positions that are exactly the same but also of
positions that are equivalent.
• •• ••• •• •••
(a)
FIGURE 7.32
(b)
(c)
(d)
We are now ready to tackle Problem 7.2 (page 214). Try it
again if you haven't already solved it. Use the various
techniques we have just discussed in order to limit the
number of cases you must consider.
We begin by simplifying the problem and considering the 2x2 version
of the game. If A takes two boxes, then B can win by taking the
remaining two boxes; if A takes one box, then B can win by taking the
diagonally opposite box—see Figure 7.33. (A can only take one of the
remaining two boxes, and so B gets the last one.) Thus, no matter what
A does, B can force a win of the 2x2 game.
FIGURE 7.33
Therefore, the position shown in Figure 7.34a is a winning position
in the 3x3 game, as is any position equivalent to it under the
symmetries of the game (that is, any position in which four boxes
remain, all of which lie in only two rows and two columns). For example,
the positions in 7.34b and 7.34c are winning positions.
Solution of
Problem 7.2

242
CHAPTER 7
• • •
• • •
• • •
(a)
FIGURE 7.34
(b)
(c)
Furthermore, the same reasoning used in the 2x2 case shows that
the positions shown in Figure 7.35 are also winning positions.
(a)
FIGURE 7.35
(b)
(c)
On the other hand, the position in Figure 7.36a and its equivalents
are losing positions, because the next player can leave Figure 7.36b
and win.
• • •
• • •
(a)
FIGURE 7.36
(b)
We saw above that there are, in essence, only three different possible
first moves for A. We will consider them now.
Case 1 If A covers three boxes as in Figure 7.28a, then B can leave
Figure 7.34a, a winning position.
Case 2 If A starts by covering two boxes. Figure 7.28b, then B can
again leave the winning position in Figure 7.34a.
Since these two openings clearly lead to a win for B, we need
consider them no further.
Case 3 A's only reasonable hope of winning is to start with one box as
in Figure 7.28c.
B now has a number of alternatives.
If she places 1 or 2 checkers, all in the same row or the same column
that A did, then we know, by virtue of deja vu, that B will lose. If B
leaves the position in Figures 7.28a or b or an equivalent, it would
be the same as if B were the first player and she started by placing 2 or
3 checkers—a losing move.

GAMES OF STRATEGY FOR TWO PLAYERS
243
Similarly, if B places 3 checkers anywhere, then A will be able to create
an equivalent of one of the positions in Figure 7.34, and thus win.
Therefore, B's only reasonable moves are:
1. Place only 1 checker, in a row and column different from the one in
which A placed his checker.
2. Place two checkers, neither of which is in the row or column in
which A placed his checker.
3. Place two checkers, exactly one of which is in the row or column in
which A placed his checker.
Because of symmetry, we need consider only the three positions in
Figure 7.37.
• •• •
(a)
FIGURE 7.37
(b)
(c)
Since the positions in Figure 7.37b and c can be obtained in one
move from the position in a, we consider b and c first. If either of these
positions leads to a win for B, then Figure 7.37a would be a losing
position since A could leave b or c. On the other hand, if b and c both
are losing positions, then we would still have to investigate position a.
Position b may be disposed of quickly, because A can win by filling
in the first column leaving Figure 7.38, an equivalent of the positions
in Figure 7.34.
We now consider position c. For each possible move that A can now
make, B has a countering move that leaves a winning position.
If A covers three squares, leaving one of the positions in Figure 7.39,
• • •
• • •
• • •
FIGURE 7.38
FIGURE 7.39 (a) (b)
then B can win by leaving the corresponding position in Figure 7.40.
• • •
• • •
FIGURE 7.40
(a)
(b)

244
CHAPTER 7
If A covers two squares, leaving one of the positions in Figure 7.41,
• • •
• • •
(a)
(b)
(c)
(d)
• • •
(e)
FIGURE 7.41
(f)
(g)
(h)
then B can leave the corresponding position in Figure 7.42.
• • •
• • •
• • •
• • •
(a)
(b)
(c)
(d)
• • •
• • •
• • •
• • •
(e)
FIGURE 7.42
(f)
(g)
(h)
(a)
FIGURE 7.43
Finally, if A covers one square, leaving one of the positions in
Figure 7.43,
• •• •• •• •• •• **
I • I I • I • I hi I • • **
I • • •
(b)
(c)
(d)
(e)
(f)

GAMES OF STRATEGY FOR TWO PLAYERS
245
then B can leave the corresponding position in Figure 7.44.
• •• •• •• ••• ••• ••
• •• •• • •• ••
• ••• • • • • •
(a)
FIGURE 7.44
(b)
(c)
(d)
(e)
(f)
In each case, B leaves a winning position.
Since position c (of Figure 7.37) is a winning position, it is no
longer necessary to consider position a.
This completes the analysis of the game. Recapping, B has a winning
strategy: If A covers two or three squares, then B covers three or two
respectively, in such a way as to leave a position that is equivalent to
the 2x2 game. If A covers one square, then B covers two squares, one
of which is in the same row or column as the square covered by A,
and the other is not. After A's next move, B will be able either to leave
a position equivalent to the 2x2 game, or to leave only two squares
uncovered, which are neither in the same row nor the same column.
1. [a| In each of the positions in Figure 7.45, it is your move. Do you PRACTICE
have a winning move? If so, what is it? PPC^RI FN^S
7.H
• • • • • • •••
• • •• •
(a)
FIGURE 7.45
(b)
(c)
(d)
2. In each of the positions in Figure 7.46, it is your move. Do you
have a winning move? If so, what is it?
• • •• *
• • •• •
(a)
FIGURE 7.46
(b)
(c)
(d)

246
CHAPTER 7
THE GAME OF NIM
While the techniques we have discussed so far in this chapter are often
helpful in analyzing games, they do not always enable us to reduce the
necessary case analysis to manageable proportions. Many games (such
as chess and checkers) are still not analyzed for this reason. Sometimes,
though, seemingly unrelated mathematical considerations lead us
through the labyrinth. One such example is an application of the binary
system (see Chapter 5) to the game of Nim, presented in Problem 7.3.
Before we present this application, we try to solve the problem with
the techniques we have discussed so far in this chapter.
If you haven't already solved Problem 7.3 (page 214),
try it again now.
Solution of
Problem 7.3(a)
As with the one pile matchstick game of Problem 7.1, we might hope
that consideration of some special cases would lead to a solution of the
general problem. We therefore begin by considering the version of the
game presented in Problem 7.3(a), in which the piles contain 3, 5, and
7 matchsticks at the beginning of the game.
Before attempting to find winning and losing positions for this
version, we investigate symmetries of the game. Since it obviously
makes no difference which pile contains 3 sticks, which contains 5, and
which contains 7, switching the order of the piles is clearly a symmetry.
Hence, there is no loss in generality in considering the pile with the
fewest sticks first and the pile with the most sticks last. Thus, we may
represent the number of sticks remaining in the three piles by an
ordered triple (a, b, c), 0 < a < b < c.
Now we may work backward to find winning and losing positions.
(Note that there are no drawing positions, since a draw is not possible.)
These are shown in Figure 7.47. Thus, (3, 5, 7) is a losing position.
It is therefore best, in the version of Problem 7.3(a), to go first and
to leave either (3, 4, 7), (3, 5, 6), or (2, 5, 7).
Solution of
Problem 7.3(b)
In the general game, we are still in the dark. The winning positions
above are still winning positions and the losing positions are still
losing positions, but it is not at all clear as to how we may generalize
to other positions. In all the winning positions above, except (3, 5, 6),
the sum of the first two numbers is the third number. We might
suspect that this relationship holds in general. That this is not the case

GAMES OF STRATEGY FOR TWO PLAYERS
247
winning
(0, 0, 0)
(0, 1, 1)
(0, 2, 2)
generalizes
previous "
steps
(0,a,a)
(1,2,3)
(1,4,5)
(2, 4, 6)
(2, 5, 7)
(3,4,7)1
(3, 5, 6)J
(0, 0, a)
(0, l,a)
(1, l,a)
(1,2,2)
(2, 2, a)
(0, 2, a)
(0,a,6)
(b, a, a)
(a, a, 6)
(1,2,a)
(1,3, a)
(2, 3, a)
(1,4,a)
(1,5,a)
(a, 4, 5)
(2,4,7)
(2, 5, 6)
(3,4,6)
(3,5,7)
osing
,a> 0
,a> ll
,a^ Ij
1
,a> 1 ^
,a>2j
,0< a< b^
,0< 6<a >
,0<a^6j
,a> 3^
,a> 3 I
,a>3j
a> 5 1
a> 5 >
, 1 ^a^3j
1
reason
objective of the game
opponent may leave (0, 0, 0)
opponent must leave (0, 0, 1)
opponent may leave (0, 1, 1)
opponent must leave (0, 1, 2) or (0, 0, 2)
opponent may leave (0, 2, 2)
opponent must leave (0, c, a), c < a
opponent may leave (0, a, a)
opponent must leave (0, a, b), a< b;
(1, l,a),a> l,or(l,2,2)
opponent may leave (1, 2, 3)
opponent must leave (0, a,b),a <b\
(l,2,a), a> 3;(l,3,a),a>3;
or (1,4, 4)
opponent may leave (1, 4, 5)
opponent must leave (0, a, 6), a< b\
(1, 2, a), a > 3; (1, 4, a), a > 5; (2, 2, a),
a > 1; (2, 3, a), a > 3;(2,4, 4); or (2, 4, 5)
opponent may leave (2, 4, 6)
opponent must leave (0, a,b),a < b\
(1,2, a), a > 3;(1, 5,7);(2, 2,a),
a > 1;(2, 3, a), a> 3; (2, 4, 5); (2, 4,7);
(2, 5, 5); or (2, 5, 6)
opponent may leave (2,5, 7)
opponent must leave one of the
losing positions listed above.
FIGURE 7.47
may be seen from the fact that (1, 3, 4), (1, 5, 6), and (2, 3, 5) are
losing positions.
Actually, the secret to generalization lies in writing all numbers in
the binary system—see Chapter 5—a point first observed by C. L.
Bouton in 1902. Consider the winning positions above written in the

248 CHAPTER 7
binary system, as in the examples in Figure 7.48. The number (written
(3,4,7) (2,5,7) (2,4,6) (3,5,6) (1,4,5)
111
1010
1101
2222
(a)
111
1010
1011
2132
(b)
FIGURE 7.50
11
100
111
222
10
101
111
222
FIGURE 7.48
10
100
110
220
11
101
LIO
222
1
100
101
202
in base ten) of I's in each column is shown below the line. In all cases,
this number is even in all columns. On the other hand, in every losing
position, at least one column sum is odd, as in Figure 7.49.
(1,3,4) (1,5,6) (2,3,5) (3,5,7)
1 1
11 101
100 110
U2 212
FIGURE 7.49
10
11
101
122
11
101
111
223
We define a position to be even if the number of I's in every column
is even (when the numbers are written in the binary system), and to
be odd otherwise. We then claim that the set of winning positions is
exactly the set of even positions, and the losing positions are the odd
positions.
To see this, we must show that conditions 1, 2, and 3 of winning
and losing positions (page 226) are satisfied.
The terminal position (0, 0, 0) is an even position, so condition 1
is satisfied.
Condition 2 says that any move made from an even position must
result in an odd position. To see that this is the case, observe that the
binary representations of two different numbers must differ in at least
one digit—one has a 1 in a position in which the other has a 0. When
a move is made from an even position, two of the binary numbers are
left unaltered, and the third is changed—at least one 0 is changed to a
1 or one 1 is changed to a 0. But then, the sum of any column in
which such a change takes place is either increased by 1 or decreased
by 1. In either case, it is changed from an even number to an odd
number, and hence the resulting position is odd.
For example, consider the even position (7, 10, 13) in Figure 7.50a.
If two sticks are taken from 13 to leave 11 = (1011)two3 the second 1
from the left in the binary representation of 13 is changed to a 0,
and the 0 is changed to 1. Hence, the second and third column sums
are changed from even to odd (Figure 7.50b).
As for condition 3, which says that from any odd position it must
be possible to reach an even position in one move, we proceed as
follows: Determine the leftmost column containing an odd number of

GAMES OF STRATEGY FOR TWO PLAYERS
249
I's. At least one of the entries in that column must be a 1 (otherwise
the number of I's in that column would be zero, an even number).
Choose any pile having a number of sticks whose binary representation
has a 1 in the column in question. Remove as many sticks from that
pile as is necessary to obtain a number whose binary representation
results in all columns having an even number of I's. For example, from
the situation in Figure 7.51a we must select the second pile, since it is
the only pile having a 1 in the leftmost odd column (the third column
from the left). The remaining two piles sum to 1201211 (Figure 7.51b),
so, to get an even number of 1 's in each column, the new number in the
second pile must have 1001011 as its representation (that is, it must be
75). This number will automatically be smaller than the number that
was originally in the second pile, because the leftmost place in which
the two numbers differ contained a 1 in the original number and a 0
in the new number. (Note that
lOllOl
1010101
1100110
2211312
101101
1100110
1201211
(a) (b)
FIGURE 7.51
2* > 2* - 1 + 2* - 2 + .. • + 2 + 1
> aA-i2*-i + a;fe_2 2*-2 + •.. 4- ai • 2 4- oo,
0 < a, < 1 for all f, regardless of which of the a, are 0 and which are 1.
See Chapter 5, page 152.)
This same consideration guarantees that it will always be possible to
replace a number having a 1 in the leftmost odd column by a smaller
number, so that all the columns contain an even number of I's.
As another example, from the position in Figure 7.52, any pile may
be chosen to work with, because all three binary representations
contain I's in the leftmost odd column. If we work with the first pile, we
must replace it by 101 to obtain an even number of I's in each column;
with the second pile, 10100 should be left; with the third pile, leave
10001.
Thus, playing Nim with three piles containing />, ^, and r sticks
respectively, we should choose to go first if the position is odd, second
if it is even. (Obviously, whether the position is odd or even will
depend on />, ^, and r.)
It should be clear that these same considerations apply even if there
are more than three piles.
1011
11010
11111
23132
FIGURE 7.52
Now that we have this simple method of determining strategy, the
Figure 7.47 chart is no longer necessary.
1. [a] Iti the game of Nim, are the following winning or losing
positions ?
(a) (3, 6, 10) (b)(3,4, 11, 12).
PRACTICE
PROBLEMS 7.1

250
CHAPTER 7
2. In the game of Nim, are the following winning or losing positions?
(a) (4, 7, 11) (b) (2, 8, 10, 14) (c) (3, 8, 12, 16).
3. [a] In each of the following positions in the game of Nim, it is
your move. You are playing against an expert. What should you do?
(a) (6, 7, 8) (b) (11, 12, 14) (c) (6, 10, 12) (d) (2, 4, 7, 11).
4. In each of the following positions in the game of Nim, it is your
move. What should you do?
(a) (3, 8, 12, 16) (b) (7, 8, 9) (c) (7, 11, 13).
PAIRING STRATEGIES
Our discussion of the game of Nim illustrated how theoretical
considerations can sometimes help in finding the winning strategy. A
particular theoretical consideration that is often helpful is the concept
of a pairing strategy. Sometimes the moves of a game or the cells of the
game board can be paired in such a way that one of the players can
always force a win by making the move that pairs with the opponent's
previous move.
For example, in the game of Problem 7.1(c) (page 214), we may think
of the moves as being paired as follows: 1 is paired with m; 2 is paired
with m — 1; 3 is paired with m — 2; and so on. (If m is odd, (m + l)/2
will be paired with itself.) Then, once a winning position is reached,
the player who reached it can win by thenceforth making the move that
is paired with the move his or her opponent makes.
As another example of a pairing strategy, consider Problem 7.4.
If you haven't already solved it, try Problem 7.4 (page 215)
again now.
Solution of
Problem 7.4
A may view the dots as being paired, as in Figure 7.53.
^ ^.^
FIGURE 7.53
If he places his first X through the middle dot, then he will be
able to force a win by always marking the dot paired with the one his

GAMES OF STRATEGY FOR TWO PLAYERS 251
opponent just marked, unless it is possible to win directly on the move.
Thus, a typical game might proceed as in Figure 7.54.
A3 B3 A4 B,
FIGURE 7.54
• X
A.
After A's first move, he plays the move paired with B's move, until
B, at her third move, leaves A the opportunity to win.
By following the above strategy, A will never leave an opportunity for
B to win. For, if A cannot win himself, then, after his turn, the left
half of the board will be symmetrical with the right half; and hence,
if no winning opportunity existed in one half, none will exist in the
other half either.
Note that the pairing in the game above is based on symmetry. This
is true of many pairing strategies. However, even in games with
symmetrical boards, it is not always true that one player can force a
win by playing symmetrically to his or her opponent. Sometimes, a
symmetry strategy will result in a quick loss.
VARIATIONS OF A GAME
Once a game is completely analyzed and a best strategy is known for
one player, X, the game becomes boring; and, of course, a
knowledgeable opponent will not want to play. (However, if X can find a
nonthinking opponent, N, he or she can easily pick up a few dollars by
baiting N—letting N win a few times and thus setting N up for the
kill.)
Sometimes, though, if a game is too difficult or too easy to analyze,
we may make some changes in the hope that a more interesting game
will result. Many different kinds of alterations are possible.
We can vary the size or shape of the "board." For example, instead
of playing the game with 28 matchsticks as described in Problem 7.1(a),
we may play the same game but start with 36 sticks. Or, instead of
having one pile of sticks, we may have several. Instead of playing Tic
Tac Toe on a 3 x 3 board, we may play it on a 4 x 4 board, or a
3x3x3 board. An interesting variation of chess is played on a
cylinder—the vertical edges of the board are bent back until they touch
each other.
In addition to varying the board, we can also vary the rules of

252
CHAPTER 7
movement. Instead of being allowed to take only 1, 2, or 3 matchsticks,
we may allow 1, 2, 3, or 4 to be taken; or we may take as many as
we like, as long as they are all in the same pile. Or we may allow a
player to make two consecutive moves, or forbid a player to make a
certain move, and so on. In a game such as chess, we may introduce
new pieces: a grasshopper (which moves along the same lines as a
queen but which, in order to move, must hop over a piece and land
on the next square) or a reflecting bishop (which may bounce off the
edges of the board as if it were a ball on a billiard table).
Still another way of varying a game is to alter the winning objective
of the game. For example, instead of trying to take the last matchstick
in Nim, you may try to force your opponent to take the last one. Or,
instead of trying to get three in a row in Tic Tac Toe, you may try
to force your opponent to get three in a row. Playing a game with
the objective of trying to force your opponent to achieve what would
normally be considered a win is usually referred to as the misere
form of the game (pronounced miz air).
The game in Problem 7.2 is a variation of Tic Tac Toe in which
both the rules for moving and the objective of the game have been
changed.
Some other variations of Tic Tac Toe will be considered in the
exercises at the end of the chapter, as will many other games and some
of their variations.
THE CHAPTER IN RETROSPECT
In this chapter you have been introduced to games in which definite
strategies are involved. These games are not only fun to play, but
analyzing games of this type further improves the ability to think
critically and solve problems.
We have also discussed some techniques for finding strategies. Some,
such as simplification, we have encountered before. Most of them you
have probably used in some form or other in everyday problem solving.
The "deja-vu" technique means, practically speaking, rely on your
previous experience in similar situations. And the symmetry technique,
again in practical terms, will remind you to recognize when two
situations are essentially equivalent.
Game analysis is closely allied to problem solving. Note again how
mathematics creeps in where you least expect it—congruences in the
matchstick game, the binary system in the game of Nim, and graph
theory in some of the exercises.
In the next chapter, we will see some one-person games in which
mathematics plays a part.

GAMES OF STRATEGY FOR TWO PLAYERS
253
Exercises
The exercises below are of two basic types. In
the first group (Exercises 7,1 through 7.72),
a game is described. Your task is to analyze
that game. Assume that two experts, A and B,
are playing the game in question; if A makes
the first move, will one player win (which one?)
or will the game end in a draw? If one player
can force a win, describe that player's winning
strategy; otherwise, describe both players'
drawing strategies. Depending on the specific
game being considered, such a description
might be a statement of some general principle
that the player should follow in selecting his
or her moves; or it might be a list of all possible
positions with which the player could be faced,
combined with a designation of what move the
player should make from each possible position ;
or perhaps just a list of winning and losing
positions is sufficient. In any case, it should be
made clear that the strategy will work no matter
what moves the opponent makes.
In many of these exercises, we examine only
some of the smaller cases of the game. A more
general discussion of most of these games may
be found in Martin Gardner's books ([18], [19],
[20], [21], [22], [23], [24], [25]) and in his
articles in Scientific American.
Boards on which many of the games discussed
can be played are illustrated on the insides of the
front and back covers of this book. For counters,
use coins or anything else your ingenuity suggests.
Be warned that we, the authors, are not
positive of the answers to all these problems.
Those about which we are unsure have been
indicated by '¥^'¥^'¥^. This does not necessarily
mean that the problem is very difficult, but
rather that we haven't had an opportunity to
do a complete case analysis.
The remaining problems in the chapter
(Exercises 7.73 to 7.83) consist of end position
problems. That is, a particular position of the
game is presented, and the challenge is to
determine what move the player who is about to
move should make to ensure a win. Or, in the
event that this player has no winning move, can
he or she assure at least a draw? In addition, you
must explain how you know that your answer is
correct.

254
CHAPTER 7
Matchstick Gaines—One Pile
Each of the following games begins with one
pile of matchsticks—or toothpicks, if you like.
The number of sticks in the pile is given in the
column headed n. If the number appearing in
this column is n itself, you should analyze the
game for all values of n in general.
The players, A and B, alternate turns. The
number of matchsticks that each player may
take at a turn is indicated in the column headed
The rule of play.
The definition of a win for the game is
indicated in the final column, headed Object
of the game. In most cases, it is either that
the player who takes the last stick wins, which
we refer to as the regular form of the game,
or it is that the player who takes the last stick
loses, which we refer to as the misere form
of the game. In the misere form, the object
is to force your opponent to take the last stick.
Note that, in problems having more than one
part (a, b, etc.) and more than one objective
(regular, misere), each objective should be
considered for each part.
Exercise
7.1.
7.2.
♦ 7.3. [a]
7.4.
♦ 7.5. [1
7.6.
♦ 7.7. [h]
n
(a) 32
[^(b)n
n
n
(a) 28
(b)«
28
60
Any odd n
The rule of play
2, 3, or 4. If only 1 stick
remains, the game is drawn.
3, 4, 5, or 6. If 1 or 2 sticks
remain, the game is drawn.
7,7 -f 1, ..., m. If only 1,
2, ..., 7 - 1 sticks remain,
the game is drawn.
1, 3, or 4.
1, 2, 3, 4, or 5; but a player
may not use the same
number his or her opponent
just used. A player who
cannot move loses.
Any perfect square
(1,4,9, 16, etc.).
1, 2, or 3.
Object of the game
(i) Regular
(ii) Misere
(i) Regular
(ii) Misere
(i) Regular
(ii) Misere
(i) Regular
[a] (ii) Misere
Regular
Regular
[a] (i) Regular: the winner
possesses an odd number of
sticks when all of the sticks
have been taken.
(ii) Misere: the loser
possesses an odd number of
sticks.

GAMES OF STRATEGY FOR TWO PLAYERS
255
Matchstick Gaines—More Than One Pile
Each of the following games begins with more
than one pile of matchsticks or toothpicks.
The number of piles and the number of sticks
in each pile is given in the column headed
k;(a^, ..., ak). For example, 3; (3,4, 5) means
that there are three piles containing 3, 4, and
5 sticks respectively.
As in the one pile games, the rule of play
column indicates how many sticks each player
may take at a turn, and the object of the game
column defines what is considered to be a win
(regular—last stick wins; misere—last stick
loses).
Exercise
^ 7.8. [a]
7.9. [h]
(Wythoff's
Nim)
7.10. [h]
7.11.11 S
7.12.
M 7.13. [h] [a]
M 7.14. [h] [a]
(Moore's Nim)
7.15.
k; (ai, ..., Gk)
2; (m, n), m < n
[a] (a) 2; (11, 15)
(b) 2; (12, 20)
3;(3, 5, 7)
3;(3, 7, 9)
(a) 3;(3, 5, 7)
(b) 3;(w, «,/))
m < n < p
(a) 4;(3, 5, 7, 9)
(b) 4; (w, «,/), q)
m <n <p < q
k; (ai, ..., a/t)
(a) 3;(3, 5, 7)
(b) 3;(m, «, p)
The rule of play
1, 2, or 3 from the
same pile.
As many as desired
from one pile, or the
same number from
both piles.
As many as desired
from one pile, or the
same number from
two piles.
As many as desired
from one pile, or the
same number from all
nonempty piles.
As many as desired
from up to two piles.
Same as Exercise 7.12.
As many as desired
from up to m piles
(m given).
Take 1 or an entire
pile.
Object of the game
(i) Regular
(ii) Misere
(i) Regular
(ii) Misere
(i) Regular
(ii) Misere
Regular
(i) Regular
[a] (ii) Misere
Regular
Regular
[a] (i) Regular
(ii) Misere

256
CHAPTER 7
Other Matchstick Gaines
7.16. [a] (Modified Kayles) Beginning with
13 matchsticks in a row as in Figure 7.55,
FIGURE 7.55
each player, at his or her turn, may take either
1 or 2 adjacent sticks. (Once a stick is removed,
the sticks to the left and to the right of the
removed sticks are not considered to be
adjacent.)
(a) [h] Analyze the regular form of this
game.
(b) Analyze the misere form of this game.
(c) [h] Beginning with n matchsticks in a
row, each player, at his or her turn, may take
either one or two adjacent sticks. Analyze the
regular form of this game.
7.17. [h] (a) Beginning with 13 matchsticks
in a circle as in Figure 7.56, each player, at
his or her turn, may take either 1 or 2 adjacent
sticks.
x^'//
FIGURE 7.56
(i) Analyze the regular form of this game.
(ii) Analyze the misere form of this game.
(b) Beginning with n matchsticks arranged
in a circle, each player, at his or her turn, may
take either one or two adjacent sticks. Analyze
the regular form of this game.
7.18. [h] Beginning with one pile containing
n matchsticks, each player must, at his or her
turn, separate any existing pile into two smaller
subpiles. The first player who cannot move is
the loser. For example, if the original pile
contains 10 sticks, A might divide it into two
piles containing 5 and 5 sticks respectively, or
3 and 7 sticks, etc. In the latter case, B might
then subdivide the 3 pile into 2 and 1, leaving
piles of 1, 2, and 7. If A then divides the 2 pile,
leaving 1, 1, 1, and 7, then B would have to
work with the pile containing 7 sticks, and
so on.
Analyze this game.
^ ^ 7.19. (Grundy's Game) Beginning with
one pile containing 14 matchsticks,
each player, at his or her turn, must separate
any existing pile into two unequal subpiles.
(Thus a pile of 4 can only be separated into
piles of 1 and 3; and a pile of 2 cannot be
subdivided.) The first player who cannot move
is the loser.
(a) [s] [a] Analyze the game.
(b) [a] Analyze the game if the pile has 15
matchsticks.
¥ 7.20. [a] Beginning with one pile
containing n matchsticks, A must subdivide the pile
into two smaller piles. B must take one of these
piles (the other pile is discarded) and subdivide
it. A then selects one of these piles (discarding
the other) and subdivides it. Etc. A player
who cannot move loses.
(a) Analyze the game if there is no
restriction as to whether or not the two subpiles are
of equal size.
(b) Analyze the game if each subdivision
must be into two unequal piles.
Related Gaines
¥¥ 7.21. [s] [a] Four aces, four 2's, four
3's, and four 4's are removed from a deck of
cards and placed face up on the table. A selects
one of the cards, turns it face down, and calls
out the face value of the card (aces count as 1).

GAMES OF STRATEGY FOR TWO PLAYERS
257
B then selects one of the remaining face up
cards, adds its face value to A's total, and
turns the card over. The face value of A's next
card is added to the previous total, and so on.
The player who reaches 22 or forces his or her
opponent to exceed 22 is the winner.
Analyze the game. ([11], Problem 476)
M 7.22. [h] [a] On a normal die, the number
1 is opposite 6, 2 is opposite 5, and 3 is
opposite 4. A places such a die on the table
and calls out the number showing on its top
face. B must then give the die a quarter turn,
so that the number previously on top is now
on one of the sides. The new top number is
added to the tot: I obtained by A, who then
gives the die a quarter turn and adds the new
top number to the previous total; and so on.
The player who reaches 26 or forces his or
her opponent to exceed 26 is the winner.
If A can originally place the die on the table
in any manner that he chooses, analyze the
game.
(Note that if, for example, a 1 is on top of
the die, then the next player may turn the
die in such a way that either 2, 3, 4, or 5
becomes the top number, but may not turn it
so that 1 remains on top or 6 becomes the
top number.) ([45], Problem 180)
7.23. Consider the following modification of
the game in Exercise 7.22. Instead of trying to
reach 26, the winning number is determined
by B at the start of the game. B may choose
any number over 20. After B chooses the
number, A makes the first move, as in Exercise
7.22.
Is there any number B can select so that she
can be sure of winning the game ?
7.24. [a] Consider the following variation of
the games in Exercises 7.22 and 7.23. A is
allowed to select the winning number but,
after he does so, B is allowed to select the
starting position of the die. A then turns the
die as before and the counting starts with the
number that is then on top.
If A is allowed to select any number over 20,
analyze this game.
7.25. [a] A is allowed to select any date of the
year other than December 31. B may then
select any date later in the same month or the
same day of any later month. For example, if
A selects June 16, then B may choose any
later date in June or the 16th of any month
from July through December. Using the same
rule with regard to the date that B has selected,
A must select a new date, and so on. The
winner is the player who arrives at December
31.
(a) What date should A select to begin with
in order to ensure the fastest possible win?
(b) What if A must start with a date in
January? ([27], May 1971)
7.26. [a] Two rooks are placed on diagonally
opposite corners of an 8 x 8 chessboard. A
moves the white rook, which is located in the
upper lefthand corner of the board, and B
moves the black rook, which begins in the
lower righthand corner. All cells that are in
the same row or column as a rook are under
attack by it. A player may move her or his
rook to any cell it attacked prior to the move,
provided that it neither passes over nor lands
upon a cell that is under attack by the op-

258
CHAPTER 7
ponent's rook. A player who cannot make a
legal move is the loser.
Analyze the game. ([13], Problem 393)
^ 7,27, [a] Consider the following variation
of the game in Exercise 7.26. A and B
command queens rather than rooks. A queen
attacks all cells that are in the row, column,
or either of the diagonal lines on which the
queen is located. A's queen starts one cell to
the right of the upper lefthand corner of the
board and B's queen begins one cell to the left
of the lower righthand corner.
Analyze the game.
^ 7.28. [h] [a] (Northcott's Nim) (a) Each
player has three counters placed on a 3 x 8
checkerboard as in Figure 7.57. At a turn,
each player may move any one of his or her
counters as far as desired to the right or left, as
long as it does not leave the same row it started
in and does not pass over or land on the piece
the opponent has in that row. The first player
who cannot move is the loser.
®
®
®
(§)!
®
®
FIGURE 7.57
Analyze this game,
(b) If the game is played on an m x n
checkerboard (w rows and n columns) with
each player having m counters, and if A starts
at the left end of each row and B starts at the
right end of each row, analyze the game.
Tic Tac Toe Variations
Each of the following games is a variation of
Tic Tac Toe. For each, the playing field, the
rules of movement, and the object of the game
are given in the chart below. As before, we
refer to the first player to move as A and to the
second player as B.
The phrase "in a row" means in cells that
are in a straight line (horizontal, vertical, or
diagonal) and contiguous. Contiguous means
that the faces, edges, or the corners of
consecutive cells are touching (see Figure 7.58 for
some possible three-in-a-row positions).
FIGURE 7.58

GAMES OF STRATEGY FOR TWO PLAYERS
259
Exercise
7.29. [s\ 0
7.30. \s\ [a]
1 7.31. H
7.32. [a]
7.33. [a]
7.34.
Size of board
3x3
3x3
3x3
4x4
3x3
4x4
The rules of play
Each player may place
either an X or an O at
each turn.
One player plays X's
and the other player
plays O's. No Player may
play in the middle or top
rows unless the X or O
placed there is supported
by either an X or O
directly below it.
Each player may place
either an X or an O,
but, to place a piece in
the middle or top rows,
it must be supported, as
in Exercise 7.30.
Same as Exercise 7.30.
One player has three X's,
the other has three O's.
Play begins as in regular
Tic Tac Toe. However,
when all six pieces are
placed, each player may,
at his or her turn, move
one of his or her pieces
one space horizontally or
vertically but not
diagonally.
Play as in regular Tic
Tac Toe (A places an X
at each turn; B places an
O).
Object of the game
(a) The player who
completes three X's or
three O's in a row is the
winner.
(b) Misere—the player
described above loses.
(a) The player to
complete three in a row
wins.
(b) Misere.
The player to complete
three X's or three O's
in a row is the winner.
The player to complete
three in a row wins.
The player to complete
three in a row is the
winner. However, if a
position is repeated
three times, then the
game is considered as a
draw.
H (a) The player to
complete three in a
row is the winner. 1
MMM (b) Misere.

260
CHAPTER 7
Exercise
7.35. [a]
7.36. [a]
M 7.37. [h] [a]
7.38. [a]
7.39. U [a]
(P. Hein's
Tac Tix)
7.40. [a]
7.41. [a]
Size of board
7x7
5x5
4x4
3x3
3x3
3x3x3
(three
dimensional)
3x3x3
The rules of play
Same as Exercise 7.34.
A places two X's, then
B places two O's, etc.
At each turn, each
player places both an X
and an O in two
horizontally or vertically
adjacent cells.
Each player at a turn
may place as many X's
as he or she desires,
provided that they are all
in one horizontal (row)
or one vertical (column)
direction. The X's placed
need not be in
contiguous cells.
As in Exercise 7.38, but
the X's must all be in
contiguous boxes in the
same horizontal or
vertical direction.
Same as Exercise 7.34.
Same as Exercise 7.40,
except that no piece may
be placed at an upper
Object of the game
The player to complete
four in a row is the
winner.
Same as Exercise 7.35.
If either player
completes three X's in a
row, then A wins. If
three O's in a row are
completed, then B wins.
If three X's and three
O's are completed
simultaneously, then the
player who made the
move completing both is
the winner.
The player who fills in
the last box loses.
(a) The player who fills
in the last box wins.
(b) Misere.
(a) The first player to
complete three in a row
is the winner.
^ ♦ (b) [s] Misere.
(a) The first player to
complete three in a row
is the winner.

GAMES OF STRATEGY FOR TWO PLAYERS
261
Exercise
7.42. [a]
7.43.
7.44.
7.45. [a]
M 7.46. [a]
7.47. [a]
7.48.
Size of board
3x3x3
3x3x3
1 X w,
n odd,
w > 3
(a) 1 X 10
(b) 1 X 12
(a) 1 X 10
(b) 1 X 9
(c) 1 X 11
1 X 16
(a) 7 X 7
(b)[A]
5x5x5
The rules of play
level unless it is supported
by a piece below it.
Same as Exercise 7.37.
Same as Exercise 7.37,
except that each
double piece placed at
an upper level must
completely rest on pieces
below it.
Each player places one
X at each turn.
Same as Exercise 7.44.
Same as Exercise 7.44.
Each player places two
X's at each turn.
Same as Exercise 7.44.
Object of the game
(b) [h] Misere.
Same as Exercise 7.37.
The player who
completes three of a kind
(regardless of whether
three X's or three O's)
in a row is the winner.
The first player to
complete three or more
in a row is the winner.
Same as Exercise 7.44.
The first player to
complete three or more
in a row is the loser.
(a) Same as Exercise
7.44.
(b) Same as Exercise
7.46.
The first player to
complete three or more
in a horizontal, vertical,
or diagonal row is the
winner, as in ordinary
Tic Tac Toe.
Dots
The game of Dots is played as follows: Given
a rectangular array (m x n) of dots, the two
players alternate turns drawing straight lines
that connect two dots which are adjacent in a
horizontal or vertical direction. Whenever a
player draws the line segment completing the
fourth side of a unit box (: :), that person
writes her or his initial in the box and must
go again, continuing the turn as long as she or
he continues to complete boxes. Note that
even if it is possible for a player to complete
a box, the player is not obliged to do so. The
winner is the player who completes the most
boxes.

262
CHAPTER 7
7.49. Analyze the game of Dots on each of
the following boards:
(a) [a] 2 X 2(: :) (b) [a] 2 x 3 (: : :)
(c) [a] 2 X 4 ♦ ♦ ♦ (d) 3 X 3
MMM (e) 3 X 4
Hex (P. Hein)
The game of Hex was described in the text
(pages 222-223).
7.50. (a) [a] [h] Analyze the game on the
following boards.
(i) 3 X 3 (ii) 4x4 (iii) 5x5
(b) Analyze the misere version (the first
player to complete a path connecting his or her
two sides of the board is the loser) on the
following boards:
^ (i) [s] 3 X 3 M (ii) [s] 4 x 4
♦ ♦ ♦ (iii) 5x5 MMM (iv) 6x6
Bridg-it (D. Gale)
The game of Gale or Bridg-it (as it became
known when it was sold commercially) is
played on a board consisting of two different
colored n x (n + I) rectangles intermeshed
with each other. For example, to construct
what we refer to as the 3x4 board, we use the
first color (say black) to make a 3 x 4
rectangular array of dots, as in Figure 7.59a, and
then use the second color (say, gray) to form
(a)
(b)
a 4 X 3 array (Figure 7.59b) which is centered
in the first array to obtain Figure 7.59c.
The players alternate turns drawing line
segments between two horizontally or
vertically (but not diagonally) adjacent dots. A uses a
black pencil to connect two black dots; then
B uses a gray pencil to connect two gray dots;
and so on. A's objective is complete a path
connecting the two black sides of the board,
and B tries to complete a path between the gray
sides. One player may not draw a line segment
that crosses a segment drawn by his or her
opponent.
7.51. [a] (a) Analyze the game of Bridg-it on
the following boards:
(i) 2 X 3 (ii) 3x4 MM (iii) 4x5
(b) Analyze the misere form (the player who
completes a path loses) on the following
boards:
(i) 2 X 3 MM (ii) 3x4
Slither (D. L. Silverman)
The game of Slither is played on a board
consisting of a rectangular array of dots. As
in the game of Dots (Exercise 7.49), players
alternate turns drawing line segments between
horizontally or vertically adjacent dots.
However, in Slither, a player's moves are restricted
by the following rule: At any time during the
game, the completed line segments must form
a continuous path that does not intersect itself.
A player who cannot make a legal move is the
loser (the winner in the misere form). For
example, in Figure 7.60a, the player who is
about to move may make any of the moves
indicated by the dotted lines.
In Figure 7.60b, the player who must move
loses because he or she has no legal move left.
FIGURE 7.59
(c)
(a)
FIGURE 7.60
(b)

GAMES OF STRATEGY FOR TWO PLAYERS
263
7.52. [h] [a] (a) Analyze the game of Slither
on the following boards:
(i) 3x3 (ii) 3x4 (iii) 4x4
(iv) 4x5 ♦♦♦(v)5x5
(b) Analyze the misere form on the
following boards:
(i) 3 X 3 (ii) 3x4 (iii) 4x4
MMM (iv) 4x5 ♦♦♦(v)5x5
Cram
The game of Cram is one in which two
players take turns placing dominoes on a
checkerboard. It is assumed that a domino is
the correct size to exactly cover two adjacent
cells of the board. Each of the following
exercises deals with a variation of this game.
Note that when we refer to an w x w board,
we mean a board having m rows and n columns.
Exercise
7.53. [a]
7.54. H [a]
7.55. H
M 7.56. [h] [a]
(Crosscram;
G. Andersson)
7.57. H
7.58. [a]
Size of board
2 X n
3x4
(a) m X n,
m odd,
n even
(b) m X n,
m even,
n even
2 X n
(a) 3 X 3
(b) 3 X 4
(c) [s] 4x4
♦ (d) 4 X 5
3x4
The rules of play
Each player places one
domino at each turn.
covering two adjacent
cells of the board.
Same as Exercise 7.53.
Same as Exercise 7.53.
Same as Exercise 7.53,
except that A can place
dominoes only
horizontally and B can
place them only
vertically.
Same as Exercise 7.56.
Same as Exercise 7.53,
except that A can place
dominoes only vertically
and B can place them
only horizontally.
Object of the game
(i) The player who
cannot move loses.
(ii) Misere form (the
player who cannot move
wins).
Same as Exercise 7.53.
Same as Exercise 7.53(i).
Same as Exercise 7.53.
Same as Exercise 7.53.
Same as Exercise 7.53.

264
CHAPTER 7
Exercise
7.59. [a]
7.60. [a]
Size of board
(a) 3 X 3
(b) 3 X 4
(c) 4 X 4
5x5
The rules of play
Same as Exercise 7.56.
Same as Exercise 7.56.
Object of the game
(i) The player who
places a domino
completing a row or
column of the board is
the winner.
(ii) Misere form (the
player who completes a
row or column is the
loser).
Same as Exercise 7.59(i).
Hexapawn (M. Gardner)
The game of Hexapawn is played on an w x w
checkerboard. The two players sit opposite
each other and each controls n pieces, called
pawns. (Assume that A controls white and B
controls black.) At the beginning of the game,
each player's pawns are placed in the n cells
of the row closest to that player. Each player
at his or her turn may move one pawn in one
of the following ways:
1. A pawn may be moved one space
vertically forward (toward the opponent's
side of the board) provided that the cell on
which the pawn will land is vacant.
2. A pawn may be moved one space
diagonally forward provided that the cell on
which the pawn will land is occupied by
an opponent's pawn. In this case, the
opponent's pawn is captured and removed
from the board.
If the cell vertically forward of the cell on
which a pawn is located is occupied and if
neither of the diagonally forward cells are
occupied by the opponent's pawns, then that
pawn may not move.
Note that even if a pawn is able to capture,
it is not obliged to do so.
The game may be won in two ways: either
one of your pawns reaches the opponent's side
of the board, or your opponent is unable to
move when it is his or her turn.
7.61. [a] Analyze the game on the following
boards:
(a) 3 X 3 (b) [h] 4 X 4
(c) 3 X w, w < 10 ♦ ♦ (d) 5 X 5
7.62. If the rules of Hexapawn above are
changed so that pawns must capture when
able to do so, analyze both the regular and
misere forms (in the latter, the player who
would win the regular form is the loser) of the
game on the following boards:
(a) 3 X 3 (b) 4 X 4 (c) 3 x w, w < 10
ffip (M. Gardner)
The game of Hip is also played on a
checkerboard. Players alternate turns placing one
checker on the board. (Assume that A places
white checkers and B places black checkers.)
The game continues until checkers of one color
complete the vertices of a square. The square
need not have horizontal and vertical sides;
the sides may be oblique, as in Figure 7.61.
In the regular version of the game, the player
to complete a square is the loser. In the misere
form, he or she is the winner.

GAMES OF STRATEGY FOR TWO PLAYERS
265
s
^ •
^'
f
V
>
FIGURE 7.61
M 7.63. [a] (i) Analyze the regular form on
the following boards:
(a) 3 X 3 ♦ ♦ ♦ (b) 4 X 4
(ii) Analyze the misere form on the
following boards:
(a) 3 X 3 (b) 5 X 5 (c) n x n for n > 5
7.64. [a] If the game of Hip is played with
both players using checkers of the same color,
analyze the misere form of the game on each
of the following boards:
(a) 3 X 3 (b) 4 X 4 (c) n x n for n odd
(d) n X n for n even
Other Checker Gaines
7.65. [a] Consider the following game which
is played on an m x w checkerboard (m rows,
n columns). The game begins when A places
a checker anywhere on the bottom row of the
board. B may then move it one space to the
right, one space to the left, or one space up
the board. Then A does the same. Etc. No
square of the board may be occupied by the
checker more than once during the game. The
game ends when the checker reaches the top
row of the board. In the regular version of the
game, the player who reaches the top row is
the winner; in the misere version, he or she is
the loser.
Analyze both (i) the regular and (ii) the
misere versions of this game on the following
boards:
(a) 5 X 5 (b) 6 X 6 (c) m x n
7.66. [h] [a] Two players alternate turns
placing checkers on a square tabletop. At each
turn, a player places one checker so that it
does not touch any of the checkers already on
the table. The player unable to place a checker
as above is the loser.
Can either player adopt a winning strategy ?
Sprouts (J. H. Conway and M. S. Pater son)
The game of Sprouts starts with n dots on a
piece of paper. Players alternate turns drawing
lines connecting two of the dots or one dot to
itself. New lines may not intersect previously
drawn lines, nor may they pass through any
dot other than the dot(s) that the line connects.
After the line is drawn, a new dot is created
somewhere on the line. A dot may have no
more than three line segments sprouting from
it (we say that a dot has three lives). If a
player cannot move when it is his or her turn,
then he or she loses.
For example, if w = 1 (there is one dot),
A has no choice but to connect it to itself. A
new dot, 2, is created (see Figure 7.62a).
Since both dots now have two segments
sprouting from them, B cannot connect either
of them to itself (this would result in four
segments sprouting from the dot). Hence, B
must connect 1 to 2. This may be done in two
ways, inside or outside (Figures 7.62b and c);
but either way, after the new point 3 is created,
1 and 2 are dead (can no longer be used) and
3 has only one life remaining (it already has
two sprouts), and so A cannot move. B is thus
the winner.
(a) (b)
FIGURE 7.62
(c)
7.67. [h] [a] (a) Analyze the game of Sprouts
for the case n = 2.
(b) Analyze the game of Sprouts for the
case w = 3.

266
CHAPTER 7
Sim (G. J. Simmons)
The game of Sim begins with n given points,
no three of which are colinear (on the same
straight hne). Players alternate turns
connecting two points by a line segment, A using a
black pencil and B using a gray pencil. Two
points may be connected, even if the line
segment between them crosses a previously
drawn line segment (but no new vertex is
created). The game ends when there is a
monochromatic triangle formed (that is, there
are three of the given points such that all three
line segments between them are of the same
color). In the regular version of the game, the
player who completes the triangle is the loser;
in the misere form, he or she is the winner.
7.68. [h] [a] (a) Analyze the game of Sim for
the following cases:
^ M (i) The regular form, for w = 5.
(ii) The misere form, for n = 5.
(iii) The misere form, for w > 5.
(b) Can the regular form, for n = 6, end in
a draw?
7.69. As a variation of Sim, consider the
game in which both players use the same color.
The game ends when a triangle is formed
(that is, all three line segments connecting
three of the given points have been drawn).
Analyze the regular and misere forms of
this game in the following cases:
(a) w = 4 ib) n = 5 M (c) n = 6
Gaines of Entrapment
In most of the games appearing so far in
this chapter, both players have the same or
similar objectives. In some games, however,
this is not the case. One class of games in
which the players have different objectives are
games of entrapment—one player tries to trap
or capture the other, while the other player
tries to escape.
One of the better known games of this type
is Fox and Hounds, also called Sheep and
Wolves. There are several variations of this
game, played on either an 8 x 8 or a 10 x 10
checkerboard. On the smaller board, the game
requires four checkers of one color to represent
the hounds, and one checker of another
color—the fox. (On the larger board, there are
five hounds.) The hounds begin the game on
the bottom row of the board and can move
only as checkers do (they use only the black
cells of the board and move forward one space
diagonally at each move). The fox starts the
game on a black cell of the top row and moves
as a king moves in checkers (it too stays on
the black cells of the board, but it moves one
space diagonally, forward or backward). There
is no jumping in this version of the game, and
so the fox can be blocked if a hound appears
in its path. The object of the hound is to
entrap the fox, so that it cannot move at all.
The fox tries to escape. It escapes if it passes
the line of hounds, because the hounds cannot
move backward.
In another version of the game, there are
twice as many hounds, and the fox can jump
over a single hound but cannot jump over two
hounds in a row.
In all versions of this game, the hounds can
win with careful play. We will not ask you to
prove this here, but we do consider a Fox and
Hounds end position problem in Exercise 7.82.
The following exercises are some related
games.
Fox and Geese
The game of Fox and Geese is closely related
to the second version of Fox and Hounds,
but it is played on a different board. The
board for Fox and Geese consists of 33 cells
in the shape of a cross, as indicated in Figure
7.63.
The X's represent geese and the O
represents the fox. The geese move one cell at a
time in a horizontal or downward vertical
direction. (They may not move vertically up-

GAMES OF STRATEGY FOR TWO PLAYERS
267
X
X
X
X
X
X
X
X
X
X
O
X
X
X
X
X
X
X
FIGURE 7.63
ward.) The fox moves one cell horizontally or
vertically (including upward). In addition, the
fox can jump over (and remove) a goose located
on a square thai the fox could move to,
provided that the following square is vacant.
The fox tries to consume the geese and to
avoid entrapment. The geese try to entrap the
fox, so that it cannot move. The geese move
first.
7.70. [h] Analyze the game of Fox and Geese.
Dwarfs and Giant
Dwarfs and Giant is another game of
entrapment, in which three dwarfs try to trap one
giant. There are several different boards on
which this game may be played. Two of these
are shown in Figure 7.64.
any cell that is numbered 5 or greater. On
board 6, the dwarfs start on I, 2, and 4, and
the giant must start on 3. At his or her turn,
the player moving the dwarfs moves one dwarf
forward or sideways from the cell in which it
is located to a vacant cell to which it is directly
connected by a line segment. Thus, for
example, a dwarf on cell 2 in board a may
move to cell 3, 4, or 5 if it is vacant; and a
dwarf on cell 2 in board b may move to 3, 5,
or 6. The moves of the giant are similar,
except that the giant may also move
backwards. The dwarfs try to entrap the giant at
the top of the board (cell 8 in a and cell 9 in 6),
occupying cells 5, 6, and 7 in a and cells 6, 7,
and 8 in b. The giant tries to avoid
entrapment, either by breaking past the dwarfs, or
by causing the exact same position to be
repeated three times (in which case the giant
is awarded the win). The dwarfs move first.
7.71.,[h] [a] (a) Analyze the game on board a
of Figure 7.64.
(b) Analyze the game on board b.
(c) What happens if the giant starts on 5, 6,
7, 8, or 9 in 6?
7.72. [h] [a] The streets of Orderly,
Maryland were planned very carefully, as the map
in Figure 7.65 shows.
and
Oblique
Street
(a)
FIGURE 7.64
On board a in the figure, the dwarfs start on
cells 1, 2, and 3, and the giant may start on
J
i."
Y
A
\
J
1st St.
2nd St.
3rd St.
4th St.
5th St.
6th St.
Ave. Ave. Ave. Ave. Ave. Ave. Ave. Ave.
ABCDEFGH
FIGURE 7.65
7th St.

268
CHAPTER 7
Slippery Sam once held up the First
National Bank of Orderly, located at the
intersection of 3rd Street and Avenue C. The
police quickly set up roadblocks at all roads
out of town. But, as they were shorthanded
that week, this left only Fred Fleetfoot, the
flatfoot, to track Sam down. Fred knew that
he must capture Sam before nightfall;
otherwise Sam would probably be able to slip
though the roadblocks. Fred succeeded in his
task.
To commemorate the occasion, the town
council manufactured the following game:
Using the diagram in Figure 7.65 as a playing
board, one player (X) plays Sam and the other
player (O) plays Fred. They start at the
locations indicated. (The Orderly police station is
located at the intersection of 6th Street and
Avenue F.) Each player, in his or her turn,
moves to a neighboring intersection. Fred's
objective is to capture (land on) Sam. Sam's
objective is to elude Fred long enough for
night to fall: If Sam has not been caught after
Fred's twenty-first move, then Sam wins.
Analyze this game if
(a) Sam goes first.
(b) Fred goes first.
End Position Problems
7.73. You are playing a game of Nim, which
began with seven piles of matchsticks. At the
moment, only five nonempty piles remain,
containing 13, 15, 18, 41, and 45 sticks
respectively.
It is your move. What should you do?
7.74. (a) You are playing the following
version of Tic Tac Toe on a 4 x 4 board. (The
3x3 version of the game was discussed in the
chapter.) Each player, at his or her turn, is
allowed to place as many X's as desired as long
as they are in the same row or column. The
player who places the final X is the winner.
At the moment, the board looks like Figure
7.66. (The letters a, b, c, and d, and the
numbers 1, 2, 3, and 4 have been added only
to facilitate reference to the individual squares.)
It is your move. What move should you
make?
a X X
b X I X I |x
c X I X I I X
d IX
12 3 4
FIGURE 7.66
(b) The board is as above, but you are
playing Tac Tix (see Exercise 7.39). That is, each
player may place as many X's as he or she
chooses, as long as they are in the same row or
column and they occupy contiguous cells.
What should your move be now?
7,75. [a] Starting with a 6 x 6 array of dots,
you and your opponent take turns placing X's
through the dots (see Exercise 7.48). Each
places one X per turn. The first to complete
three or more adjacent X's in a horizontal,
vertical, or diagonal row as in Tic Tac Toe
is the winner.
It is your turn to move from the position in
Figure 7.67.
Where should you place your next X?
b •
c •
d •
e •
f •
12 3 4
FIGURE 7.67
5 6
7.76. [h] You are playing Dots (see Exercise
7.49) on a 4 X 4 board. The position is as
shown in Figure 7.68 and it is your move.

GAMES OF STRATEGY FOR TWO PLAYERS
269
A B C D
-•F *G jH
i| tj Ik Jl
M N O P
FIGURE 7.68
What should you do to ensure a win?
(Recall that a player is not obligated to
complete a box if it is possible to do so, but that,
if a player does complete a box, then he or
she must make another move.)
7.77. In a game of Hex on a 6 x 6 board,
you are White. The position is as shown in
Figure 7.69.
white
black
It is your move. What should you do ? And
how should the play continue?
7.78. In a game of Bridg-it (see Exercise 7.51)
on a 5 X 6 board, you are gray (vertical). The
position is as shown in Figure 7.70 and it is
your move. What should you do, and how will
the game proceed from there?
a • • • • •
b .1. — . .
C i • I • f
e • y ^ * ^ f i
i • 4 • • •
k 1 • • • •
12 3 4 5 6 7 891011
FIGURE 7.70
7.79. 0 In a game of Slither (see Exercise
7.52) on a 6 x 6 board, the position is as shown
in Figure 7.71.
• ^
• • • • •
1 n. i
FIGURE 7.71
5 6
It is your move. What move should you
make? How should you continue to play
thereafter ?
7.80. You are White in a game of Hexapawn
(see Exercise 7.61) on an 8 x 8 board. The
position is as shown in Figure 7.72.
a
b
c
d
e
f
g
h
WP
BP
BP
BP
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
12 3 4
FIGURE 7.72

270
CHAPTER 7
The outcome appears bleak. But it is your
move and, by making the correct move, you
can actually win. What move should you
make?
7.81. In a rematch of Hexapawn, on a 9 x 9
board, you reach the position shown in Figure
7.73.
a
b
c
d
e
f
g
h
i
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
BP
WP
123456789
FIGURE 7.73
You are White and it is your move. What
should you do?
7.82. [a] You are playing Fox and Hounds
(see Games of Entrapment, page 266) on an
8x8 board. It is your move. The position is
as shown in Figure 7.74.
a
b
c
d
e
f
g
h
H
F
H
H
H
12 3 4 5 6 7
FIGURE 7.74
(a) What should you do if you are the fox ?
(b) What should you do if you are the
hounds?
7.83. You are playing a game of three
dimensional Tic Tac Toe on a 4x4x4 board
(see Figure 7.75).
3/
a
b
C
(
^A / / /
2/ / / /
' /i/ /
V / X /
,--'-
"'";
f
y
■y-
1
)- -
•"".
"';
<
/l
\
/
3
J
A
I
level IV
level III
level II
level I
abed
FIGURE 7.75
To facihtate playing the game, the board
may be visualized as in Figure 7.76.
Your objective is to complete four O's in a
straight line. The line may lie in a horizontal,
vertical, or any of several diagonal directions.
Your opponent's objective is to complete four
X's in a straight line.
In visualizing the board as in Figure 7.76,
you must be careful to note whether or not a
set of X's or O's constitutes a win. For
example, the four X's in Figure 7.77 are
actually in a straight line as are the four O's
(see Figure 7.78).

GAMES OF STRATEGY FOR TWO PLAYERS
271
1
2
3
4
level I
(bottom)
FIGURE 7.76
1
2
3
4
a b c
level II
1
2
3
4
a b c
level III
1
2
3
4
level IV
(top)
X
ol
level I
(bottom)
FIGURE 7.77
O
X
abed
level II
X O
abed
level III
X
o
a b e
level IV
(top)
In the game you are currently playing, you
are O, and the position shown in Figure 7.79
(or, equivalently. Figure 7.80) has been
reached. It is your move.
How can you be sure of winning the game ?
1
3/
Y
\ /
''
\/C
a
b
e
A / / /c
2/^y / /
' /\/ /
/ ;< /
/
"/'
^
^
/
/
y-
-4-^
y^
'Y
^^-
-T -^
y
u -
/
y
^/C
__zl
/A
^
/
_^^
/
'->}'
-7
/-
/a
v^-
/
^
d
A
/I
/I
A
abed
FIGURE 7.78
level IV
level III
level II
level I

272
abed
yy//
4>9^
k<^
'^^'A.'
K'.
'/Ti
'A-
1^
q>/
IQ
k^i
abed
FIGURE 7.79
level IV
level III
level II
level I
CHAPTER 7
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
X
X
X
abed
O
O
X
X
o
o
o
X
X
o
o
o
X
o
X
abed
O
o
o
X
X
o
o
abed
X
X
X
o
X
abed
FIGURE 7.80
level IV
(top)
level III
level II
level I
(bottom)

8
Solitaire Games
and Puzzles
In Chapter 7, we considered two person games—games in which
each player competes against an opponent. In this chapter, we consider
" solitaire" or one person games, many of which are well known and of
ancient origin. In a sense, all recreational problems are one person
games; however, here we are primarily interested in games where
counters or pieces are moved or manipulated.
With most puzzles of this type, finding the solution appears at first
to be just a matter of trial and error; but, oftentimes, mathematics
plays an important role in the ultimate solution.
As usual, the chapter begins with a number of sample problems.
Some of these may be familiar to you because versions of them have
been sold commercially. Try them, by trial and error if necessary;
models for them can easily be made out of paper. If you are unable
to obtain a systematic solution, do not be discouraged. You will get a
feeling for the game just by playing it several times.
In many cases, the mathematical ideas necessary for the solutions
will seem at first to bear no relationship to the problem; you may
feel as if the ideas have been pulled out of a hat. In a sense, this is
the reason why these particular problems were chosen. It's exciting to
see how mathematics can be applied in unlikely looking situations. The
usefulness of seemingly unrelated ideas is an important facet of problem
solving in general, in all fields of endeavor. Many of the best problem
solvers are people who are able to go beyond the apparent confines of
a problem and draw on knowledge and ideas from other areas.
273

274
CHAPTER 8
Problem 8.1
A popular puzzle sold in many stores consists of a wooden board
supporting three posts, on one of which is found a set of seven rings
of diminishing diameters, with the largest ring on the bottom and the
smallest on the top. The object of the puzzle is to transfer all seven
rings from the post they are on to a different post, subject to the
following rules:
SAMPLE
PROBLEMS
o o o
o o o
ooooooo
ooooooo
ooooooo
coo
GOO
FIGURE 8.1
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
FIGURE 8.2
1. Only one ring may be moved at a time;
thus, a move consists of taking the
uppermost ring from one post and
placing it on another post.
2. At no time may a larger ring be placed on top of a smaller ring.
(a) What is the smallest number of moves in which the required
transfer can be effected?
(b) If, instead of seven rings, there are n rings, what is the smallest
possible number of moves (expressed as a function of n) in which the
transfer may be carried out?
Problem 8.2
(a) Assuming that a domino is exactly big enough to cover two
neighboring cells of a checkerboard, can an ordinary 8x8
checkerboard from which two diagonally opposite corner cells have been
removed be covered by a set of 31 dominoes ?
(b) Is it possible to cover all but one cell of an 8 x 8 checkerboard
by using 21 straight trominoes? A straight tromino looks like [ 1 Hi?
and is assumed to be just the right size to cover three cells of the
checkerboard. ([26], pp. 20-21)
Problem 8.3
Consider a board containing 33 holes, as shown in Figure 8.1. There
is a peg in each hole except the center hole.
Whenever it occurs that two adjacent (horizontally or vertically)
holes are occupied and that the next hole in that same line is empty,
the two pegs from the occupied holes may be removed and one of
them then placed in the third hole. (In other words, one peg jumps
over the other, landing in the vacant hole; the peg that was jumped is
removed.)
The object of the game is to leave only one remaining peg. When
this is accomplished, in which holes could the final peg possibly lie?
Problem 8.4
Consider a flat box containing the numbers 1 through 15 arranged as
in Figure 8.2.

SOLITAIRE GAMES AND PUZZLES
275
A move consists of sliding any number adjacent to the empty space
(in the diagram above, 12 or 15) into that space, thereby creating
a new arrangement of the numbers.
Find a sequence of moves—if it is possible to do so—starting with
the position shown in Figure 8.2 and resulting in
(a) the arrangement shown in Figure 8.3a.
(b) the arrangement shown in Figure 8.3b.
Problem 8.5
Given four cubes with faces colored as in Figure 8.4, arrange them
in a column so that each side (front, back, left, and right) of the stack
shows each of the four colors (red, white, blue, and green) exactly once.
cube I:
/
w
[15
11
7
3
14
10
6
2
13
9
5
1
12]
8
4
(a)
12
8
1 4
15
11
7
3
14
10
6
2
13
9
5
1
(b)
FIGURE 8.3
W
W
W
cube II:
/
W
W
cube III:
W
/
W
w-
w
cube IV:
W
/
B B
\ B
W
I
G
FIGURE 8.4

276 CHAPTER 8
The problems considered in this chapter are of two basic types.
We are asked either to perform some particular manipulative task or
to determine exactly which manipulative tasks are possible to perform.
In the latter case, we frequently use some form of mathematical
reasoning to limit possibilities—to show that certain tasks are
impossible. Thus, we know what we cannot do, but not what we can
do. We are then left with the problem of showing that the tasks which
remain are actually possible. This frequently reduces to a problem of
the first type—we can show that a task is possible by actually carrying
it out.
A frequent question associated with problems of the first type is
how to perform the required manipulation in the minimum number
of moves. This question is often very difficult to answer; but, in some
puzzles of this type, mathematics can again come to our aid.
Problem 8.1 is such a puzzle, and there is a story* that goes with it.
THE TOWER OF BRAHMA
In the great temple at Benares, beneath the dome which
marks the center of the world, rests a brass plate in which
are fixed three diamond needles, each a cubit high and as
thick as the body of a bee. On one of these needles, at the
creation, God placed sixty-four discs of pure gold, the largest
disc resting on the brass plate, and the others getting smaller
and smaller up to the top one. This is the tower of Brahma.
Day and night unceasingly the priests transfer the discs from
one diamond needle to another according to the fixed and
immutable laws of Brahma, which require that the priest on
duty must not move more than one disc at a time and that
he must place this disc on a needle so that there is no smaller
disc below it. When the sixty-four discs shall have been thus
transferred from the needle on which at the creation God
placed them to one of the other needles, tower, temple, and
Brahmans alike will crumble into dust, and with a
thunderclap the world will vanish.
As a result of this story, the game described in Problem 8.1 has
become known as the Tower of Brahma. It is also known as the
Tower of Hanoi, although we are not sure why.
Before reading the solution of Problem 8.1 (page 274), you
should try the problem again if you haven't already solved it.
* The story is taken from Ball [3], p. 304. He attributes the story to De Parville at
about the time that the puzzle was brought out by E. Lucas (c. 1883).

SOLITAIRE GAMES AND PUZZLES
277
We now try to solve Problem 8.1(a).
After some trials, one can see that, before the seventh ring can be
moved, the top six rings must first be moved and must be in the
position shown in Figure 8.5. (If any other ring were still on A, the
Solution of
Problem 8.1(a)
II
FIGURE 8.5
seventh ring could not be moved; and if there were rings on both B
and C, then the seventh—the largest—ring would have no place to go.)
At this point, the seventh ring is moved to peg B; and then the six
rings, step by step, must be moved so that they end up on top of
ring 7 on peg B.
Thus, if
m-f
and
then
the least number of moves required
to move the pile of seven rings,
the least number of moves required
to move a pile of six rings.
+
1
to move 6 rings to move ring 7 to move 6 rings
to peg C to peg B back to peg B
= 2m^ + 1.
The problem now reduces to a similar one tor fewer rings. If
then
and so
the least number of moves required
to move a pile of five rings.
m^ = 2m^ -f 1
rrij = 2(2ws + 1) + 1 = 2^m^ 4-2^+1.

278
CHAPTER 8
Continuing this way,
m, = 2\2m^ + 1) + 2^ + 1 =
= 2X + 2^ + 2^ + 1 = • • •
= 2^m, + 2^ + 2^* + 2^ + 2^ + 2^ + 1
= 2^ + 2^ + 2^ + 2^ + 2^ + 2^ + 1 (m^ = 1)
= 2^-1 (as we saw in Chapter 5, page 152)
= 127.
Solution of Similarly, it can be verified, for w = 1, 2, 3, 4, ..., 7, that
Problem 8.1(b)
n
1
2
3
4
7
f^n
1 = 2* - 1
3 = 2^ - 1
7 = 2^ - 1
15 = 2" - 1
127 = 2' - 1
Based on the pattern in the above chart, our guess is that, for Problem
8.1(b), the least number of moves required to move a pile of n rings is
m„ = 2" - 1. This is only a conjecture, however, and needs proof. We
use the method of mathematical induction.*
Let S„ be the statement that the least number of moves in which a
pile of n rings can be moved from one peg to another is m„ = 2" — 1.
Si is true, as one ring may be moved in 2^ — 1 = 1 move.
Assume that S„ is true. That is, assume m„ = 2" — 1.
To move n + \ rings, we must first move n rings from A to C, then
move the (n + l)st ring from A to B, and finally move the n rings
from C to A. Thus
m„ + I = 2mn + 1
= 2(2" - 1) + 1
= 2"+! - 1.
But this is statement »S„ + i. Thus, if Sn is true, so is »S„ + i.
* This topic is discussed in Appendix B. Mathematical induction has been mentioned
before in this book and we hope you have been curious enough to read this appendix.
In any case, we finally carry out an inductive proof in detail.

SOLITAIRE GAMES AND PUZZLES
279
Since S^ is true, the inductive argument implies that S2 is true;
therefore, the inductive argument implies that S3 is true; etc. Thus,
S„ is true for every positive integer n. This completes the proof.
By mathematical induction, then, the minimum number of moves
required to move a pile of n rings is 2" — 1.
When, then, will the world come to end? Assuming that the
Brahmans move one ring per second and that they never rest, the
world will come to an end 2^"* - 1 seconds after it was created. But
you need not rush to get your affairs in order; 2^** - 1 seconds is over
500 billion years.
DISSECTION PROBLEMS
Another type of solitaire puzzle of ancient origin involves geometrical
dissection and reassembly. In this type, you cut up a given geometrical
figure and reassemble the pieces to form some other specified
geometrical figure. One of the best known and most fundamental theorems
of mathematics, the Pythagorean Theorem, was probably first proved
c. 550 BCE using a dissection proof—although it was known by the
ancient Chinese (c. 1100 bce). The theorem says that the sum of the
squares of the two legs of a right triangle is the square of the
hypotenuse—that is, if a and b are the lengths of the legs of a right
triangle and if c is the length of the hypotenuse, then c^ = a^ -\- b^.
The proof in question proceeds by dissecting the square of side a -t- b
in two different ways. In the first way, the square is broken into four
right triangles, congruent to the given one, and two squares of sides a
and b respectively. The second dissection breaks the square of side
a + b into four right triangles, congruent to the given one, and one
square of side c (see Figure 8.6). Since the four triangles in Figure
8.6a are all congruent to, and hence cover the same total area as the
four triangles in Figure 8.6b, the remaining two squares in Figure 8.6a
must cover the same area as the remaining square in Figure 8.6b—
^2 _|_ ^2 _ ^2^ gg required.
b a
a\ ^\^
b\ /
Pythagorean Theorem:
FIGURE 8.6
(a)
(b)

280
CHAPTER 8
FIGURE 8.7
There are many different proofs of the Pythagorean Theorem,
including a number of other proofs by dissection [see Exercise 8.5(c)].
A much deeper theorem of geometry says that if A and B are any
two planar polygonal figures (figures with straight line segments as
edges) having the same areas, then it is possible to dissect A into a
finite number of pieces that can then be reassembled to form B. For
example, a square may be cut into four pieces that can be reassembled
to form the letter T (see Figure 8.7).
Most recreational problems of this type ask not only that we dissect
and reassemble, but that we use the fewest cuts possible in doing so.
Many such problems are still unsolved in the sense that, although a
dissection and reassembly have been found that appear to make use of
a minimal number of cuts, no one has actually been able to prove that
fewer cuts will not suffice. A description of some of the techniques used
in solving dissection problems may be found in [37]. In general, the
solution of these problems seems to require considerable trial and error
and a great deal of patience.
In some sense, the discussion up to this point does not seem to belong
in this chapter, for the problems discussed so far are not really
manipulative—at least, not until the dissection part of the problem has
been completed. In many problems, however, we are given the pieces
that we are to use, and the problem is to assemble them.
Probably the best known recreations of this type, purported by some
to date back over 4000 years, are Tangrams. Traditionally, a square
is cut up into pieces as indicated in Figure 8.8 (other ways of cutting
Tangram square
FIGURE 8.8
a square or even a rectangle have also been used occasionally), and the
object is to combine all of the pieces to make various pictures. For
example, a shark may be made as in Figure 8.9.

SOLITAIRE GAMES AND PUZZLES
281
shark
FIGURE 8.9*
Sometimes more than one set of Tangrams are combined to make
more intricate pictures. For example, a microscope can be made using
two Tangram sets, as Figure 8.10 shows (at a decreased scale).
The solutions of problems such as these require measurement of
line segments and angles, but the solutions also seem to depend,
to a large extent, on trial and error. For this reason, we do not dwell
on problems of this type, but instead turn our attention to related
problems whose solutions depend more heavily on reasoning.
POLYOMINOES
In what follows, we are primarily interested in dissecting a checkerboard
or modified checkerboard into pieces that are made up of a fixed
number of cells (squares). (Much of the theory in this area has been
developed by S. Golomb and may be found in his book [26].) Pieces
that contain two cells look like fT I and are called dominoes, because
of their resemblance to ordinary dominoes. Pieces with three cells are
called trominoes; pieces with four cells are called tetrominoes; pieces
with five cells are called pentominoes; and so on. Although all dominoes
look alike (f I 1 or , which are essentially the same), there are actually
two distinct kinds of trominoes—the straight tromino and the right
tromino, shown in Figure 8.11.
Similarly, there are five or seven tetrominoes—see Figure 8.12—
depending on whether d and f (and similarly, e and g) of the figure are
FIGURE 8.10*
FIGURE 8.11
* Based on [53], Problems 84 and 322.

282
CHAPTER 8
(a)
(b)
(c)
(d)
(e)
(0
(g)
FIGURE 8.12
considered as the same or different. The only way that f can be
obtained from d is by flipping the piece over so that the face that was
originally facing up ends up facing down. If the front and back of a
tetromino are indistinguishable from each other, then this flipping is
permissible and the two pieces would be considered as being the same;
however, if the front and back differ (for example, if they are
different colors), then the two pieces must be considered to be
different. Unless otherwise indicated, tetrominoes and other poly-
ominoes here are assumed to be reversible, and so we consider only the
five tetrominoes shown in Figure 8.12a, b, c, d, and e.
It is easily seen that a set of 32 dominoes can cover a standard
8x8 checkerboard; or, equivalently, that a standard checkerboard can
be dissected into a set of 32 dominoes. Similarly, an w x w checkerboard
can be dissected into a set of ww/2 dominoes, if either w or w is even.
If both m and n are odd, then obviously an m x n board cannot be
covered by dominoes, since each domino covers an even number of
cells (2) and the board contains an odd number of cells. However, the
remaining board can be covered with dominoes, if an appropriate cell
is removed.
We are now ready to solve Problem 8.2(a), page 274.
Try it again if you haven't already solved it.
Solution of
Problem 8.2(a)
In this problem, two cells have been removed from an ordinary 8x8
board. Because an even number (62) of cells remain, our first impulse
is to think that the required covering can be done. However, after
trying for a while to do it, we begin to suspect that maybe it cannot
be done after all. The key to the solution of the problem is to realize
that both cells that have been removed from the board are of the same
color. Thus, 32 black but only 30 white cells remain. A domino must

SOLITAIRE GAMES AND PUZZLES
283
necessarily cover one cell of each color. Thus, 31 dominoes must cover
31 black cells and 31 white cells. Therefore, 31 dominoes cannot cover
the given board.
On the other hand, if any two cells of opposite color are removed,
then the remaining board can be covered [Exercise 8.9(a)].
The idea of coloring plays a role in many other problems,
especially to show that certain positions cannot be obtained. As an
example, we consider Problem 8.2(b).
We again ask you to try Problem 8.2(b), page 274, if
you haven't already solved it.
This time, since we are interested in trominoes, we recolor the board Solution of
in three colors, so that any straight tromino placed on the board must Pfohl^m fi ^^h"^
necessarily cover one cell of each color (see Figure 8.13). • \ /
FIGURE 8.13
A set of 21 trominoes must therefore cover 21 cells of each color.
Because there are 22 black cells and only 21 of each of the other colors,
the cell that will not be covered must be one of the black ones. In
other words, all of the cells colored gray or white in the Figure 8.13
board must be covered. Note, however, that this coloring is somewhat
arbitrary. Any other coloring scheme in which every tromino must
cover one cell of each color could also be used. Applying any
symmetry of this board (see Chapter 7, page 235) produces another
coloring having 22 black cells and 21 cells of each of the other colors.
In particular, using vertical reflection, we get Figure 8.14. Again, the
uncovered cell must be one of the black ones. The only cells that
are black in both colorings are those shown in Figure 8.15. These

284
CHAPTER 8
FIGURE 8.14
FIGURE 8.15
are equivalent to each other under all symmetries of the board, and
so we may select any one of them to be the cell that remains
uncovered.
Now rhar we know which cell to leave uncovered, we must find a
covering that works. One such covering is shown in Figure 8.16.
FIGURE 8.16
Many other problems of polyominoes have been investigated (see
[26], for example). In particular, pentominoes have generated much
interest. Some questions relating to these and the other polyominoes
may be found in Exercise 8.9 at the end of the chapter.
SOMA
It is natural to attempt to generalize the ideas presented so far to
higher dimensions. Instead of considering planar regions made up of
a number of connected squares, we might investigate three dimensional
figures made up of a number of connected cubes.

SOLITAIRE GAMES AND PUZZLES
285
As with squares, there is only one way of sticking two cubes
together, and there are two ways of attaching three cubes. However,
the extra dimension enables us to increase the number of ways in
which four cubes can be connected. Considering two shapes obtained
by connecting the cubes to be the same if and only if one can be
turned in 3-space to obtain the other, we obtain the eight different
possibilities shown in Figure 8.17.
3
^^^
(1)
(2)
(3)
(4)
F
/■
/ /
IGL
JRE
~~7~
/
/
7\
V
8.17
(6)
(7)
(8)
Note that, although (7) and (8) are very similar, they are different
because, no matter how (7) is turned, it cannot be made to look exactly
like (8).
Piet Hein, whom we already met in connection with the game of
Hex, observed that the last six of the solids pictured in the preceding
figure may be combined with the tricube in Figure 8.18 to form a
3x3x3 cube. In fact, this may be done in many different ways. This
observation led Hein to invent the puzzle Soma, the object of which
is to use the seven pieces—(3), (4), (5), (6), (7), (8), and (9)—to form
various three dimensional shapes. Coloring can sometimes be useful
here too to show that certain configurations are not possible. (For
example, see Exercise 8.10.)
(9)
FIGURE 8.18
In the solution of the tromino problem—Problem 8.2(b)—we
colored a grid with three colors to determine which square could be
removed from a standard checkerboard so that the remaining board
could be covered with straight trominoes. The same basic idea is also
useful in the solution of Problem 8.3.
PEG SOLITAIRE
The game of Peg Solitaire dates back at least as far as the early 1700s
and is probably much older. Through the ages, it has been played on

286
CHAPTER 8
boards of many different sizes and shapes, but there are only two
basic variations of the game.
In the first version, a move consists of one peg jumping over an
adjacent peg and landing in a vacant hole on the other side of the
jumped peg. The jumped peg is then removed from the board. The
object of the game is to leave one peg (or a specified number of pegs)
remaining.
In the second version, the object is to move a set of pieces from
the given starting position to a given terminal position, or to interchange
the positions of two sets of pieces of different colors. A piece may be
moved from the square (hole) it occupies to any adjacent vacant square,
or a piece may jump, as in the first version, except that the jumped
piece is not removed from the board. (See Exercise 8.12 for problems
of this type.)
The game in Problem 8.3 obviously falls into the first category, and
is played on a cross-shaped board. This version of the game has been
sold commercially under the name Hi Q.
If you haven't already solved Problem 8.3 (page 274),
try it again now.
Solution of We can think of the peg board as having squares rather than holes.
prpvk^l^ppj ft ^ If we then color the board as in the tromino problem, we obtain
Figure 8.19.
n
n
J
FIGURE 8.19
The key to the solution of Problem 8.3 is the observation that any
move in the game involves one cell of each color. Two of the three
cells involved were originally occupied but end up vacant, and the
third cell was originally vacant but ends up occupied. Thus, if G, B,
and W respectively denote the number of gray, black, and white cells
that are occupied by pegs, then any move changes G, B, and W by 1
(two of them decrease by 1 and the third increases by 1).

SOLITAIRE GAMES AND PUZZLES
287
But this means that the relative parities of G, B, and W are not
changed by any move. That is, if G and B were originally of the
same parity—both were odd or both were even—then they are still of
the same parity after any move. (The parity of each is changed, and
therefore they remain of the same parity as each other.) Similarly, if G
and B were originally of opposite parities—one odd and one even-
then they remain of opposite parity after any move. A similar discussion
applies to G and W, and to B and W.
In the starting position of Problem 8.3, using the coloring indicated
above, we obtain Figure 8.20. Note that G and W are of the same
parity and B is of the opposite parity.
In a final position, with only one peg remaining, we must have one
of the following:
G-1, B = W = 0; B=l, G = W = 0; W=l, G = B = 0.
But G and W must be of the same parity after each move, since they
are of the same parity to begin with. Thus, the only possibility is B = 1,
G = W = 0; and so the one remaining peg must end up on a black
square.
Applying this same reasoning to the symmetrical pattern (Figure
8.21), we find that the last peg must again be on a black square. The
only squares that are black in both coloring schemes are shown in
Figure 8.22.
G =
B =
W =
FIGURE 8.20
m
M
a
FIGURE 8.21
FIGURE 8.22
To see that it is possible to have the final peg remain in any of
these five squares, it is only necessary to find a sequence of jumps that
accomplishes the desired task. We know of no algorithm or theoretical
approach for finding such a sequence; rather, it seems to be a matter of
reasoned trial and error, and of patience. To present one sequence
that works, we number the cells of the board as in Figure 8.23. Note
that the first digit in each box represents the row number (from top
to bottom) and the second digit represents the column number (from
left to right).
31
41
51
32
42
52
13
23
33
43
53
63
73
14
24
34
44
54
64
74
15
25
35
45
55
65
75
36
46
56
37
47
57
FIGURE 8.23

288
CHAPTER 8
A jump will be denoted by the number of the cell in which the
jumping piece starts followed by the number of the cell where it lands.
For example, 46-44 means that the piece in 46 jumps over the piece in
45 and lands in 44.
The following sequence of moves leaves one piece in 44:46-44, 65-45,
57-55, 54-56, 52-54, 73-53, 43-63, 75-73, 73-53, 35-55, 15-35,
23-43, 43-63, 63-65, 65-45, 45-25, 37-57, 57-55, 55-53, 31-33,
34-32, 51-31, 31-33, 13-15, 15-35, 36-34, 34-32, 32-52, 52-54,
54-34, 24-44.
Just prior to the last move in the sequence above, there are two
pegs left—one in 24 and one in 34. If the last move had been 34-14
instead of 24-44, then a lone peg would be left in 14.
To obtain a single peg in 47, 74, or 41, we need only apply
rotational symmetry to the sequence that leaves a single peg in 14.
Note that, since we start with 32 pegs and end with 1 peg, a total
of 31 jumps must be made. However, if we count successive jumps by
the same peg as part of the same move, it becomes meaningful to ask
" What is the smallest number of moves in which one peg can be left? "
For several years, it was thought that the answer was 19, until E.
Bergholt (in 1912) presented the 18 move solution given above, and
claimed that no one would ever beat it. It was not until recently, though,
that it has actually been proven that no solution with fewer moves is
possible. (If a different space is left vacant in the starting position, then
it is possible to leave a lone peg in fewer than 18 moves, but the
remaining peg will not end up in 44.) In general, questions of minimality
are very difficult, because theoretical considerations seem to be of little
help.
THE FIFTEEN PUZZLE
The ideas of coloring a grid or checkerboard and of parity can also
be used in the solution of yet another puzzle, which was first introduced
in the 1870s by Sam Loyd, one of the greatest American problematists.
The puzzle was originally entitled the Boss Puzzle, but has become
known as the Fifteen Puzzle for reasons that will soon be obvious.
The game consists of a flat box containing 15 movable pieces,
numbered 1 through 15. To begin with, the pieces are arranged with
a blank space in the lower righthand corner, as in Figure 8.24.
The object is to arrive at different specified arrangements of the digits
by sliding the pieces about. Any piece adjacent to the blank space
may be slid into the blank space, thereby creating a new blank
space and a new arrangement. No piece may be removed from the
box at any time.
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
FIGURE 8.24

SOLITAIRE GAMES AND PUZZLES
289
Since there is only one blank space, a move may be described by
indicating the number of the piece being moved. Thus, 12, 11, 7
indicates the sequence of moves in Figure 8.25.
FIGURE 8.25
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
12
->
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
11
->
1
5
9
13
2
6
10
14
3
7
15
4
8
11
12
7
1
5
9
13
2
6
10
14
3
7
15
4 1
8
11
12
As originally proposed by Loyd, the problem was to begin with the
starting arrangement indicated in Figure 8.24 and to obtain the
arrangement in Figure 8.26. He offered a $1000 prize to the first person
submitting a correct solution. Although many people attempted this
problem, Loyd knew that his money was safe—the problem is
impossible to solve.
How does one show that the task in question is impossible? And how
does one know what arrangements are possible? As with Hi Q, we will
first use theoretical considerations to show that certain positions are
impossible to obtain and then we will show how all others can be
obtained. We present two approaches to the solution of this problem.
The first makes use of parity arguments and the second combines
parity with coloring.
Let us first, though, simplify the puzzle and consider a 2 x 2 version
of the game. We start with Figure 8.27 and ask what arrangements of
the pieces may be obtained by following the above rules.
The only moves we have available are to rotate the pieces clockwise
or counterclockwise. There are only twelve arrangements possible, all
obtainable by clockwise rotation. These are shown in Figure 8.28.
\l
5
9
13
2
6
10
15
3
7
11
14
4
8
12
FIGURE 8.26
1
3
2 ■
FIGURE 8.27
1
3
2
2
3
1
1
3
2
2
3
1
I 3
1
2
2
1
3
3
1
2
2
1
3
3
2
1
1
2
3
3
2
1
1
2
3
FIGURE 8.28
We are essentially dealing with arrangements of the integers 1, 2,
and 3 within the 2x2 box. We can associate a triple (a, b, c) with
each arrangement as follows: List the integers in the order given by the
pattern in Figure 8.29, ignoring the blank. For example, the arrange-
^
FIGURE 8.29

290
CHAPTER 8
1
2
3
and
1
3
2
FIGURE 8.30
2
1
3
FIGURE 8.31
-<
-^
y^
FIGURE 8.33
1
3
2
FIGURE 8.34
ments in Figure 8.30 both correspond to (1, 2, 3), and that in Figure
8.31 corresponds to (2, 3, 1), etc.
Any ordering of the numbers 1, 2, and 3 (or, in general, 1, 2, ..., w)
is called a permutation. By the Multiplication Principle (see Chapter
1, page 19), there are n(n — l)(w — 2) • ... • 2 • 1 distinct permutations
of the n integers 1, 2, ..., w. For w = 3, the 3 • 2 • 1 = 6 permutations
are: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1).
The permutations (3, 1, 2), (2, 3, 1), or (1, 2, 3) all correspond to
arrangements that could be obtained (see Figure 8.28) in the 2x2
game starting with Figure 8.27.
There are three other possible permutations of the numbers 1, 2,
and 3: (3, 2, 1), (2, 1, 3), and (1, 3, 2). These respectively correspond
to the arrangements in Figure 8.32 (among others), which are not
attainable from the starting position in Figure 8.27.
3
1
2
FIGURE 8.32
2
3
1
1
2
3
Note that the permutation we associate with a given arrangement
depends on the pattern or listing order that we use. We have chosen
this particular listing in the 2x2 case because the permutations that
arise distinguish between obtainable arrangements and those that are
not obtainable. If, instead of using the pattern in Figure 8.29, we
list numbers in a different order—for example, upper left corner, upper
right corner, lower left corner, lower right corner (see Figure 8.33),
then we could get all six possible permutations. For example, using the
new listing order, the attainable arrangement in Figure 8.34 would give
rise to the permutation (1, 3, 2).
In this case, considering permutations would be of no value to us
because the method would not eliminate any permutations and so we
would not be able to say that only arrangements corresponding to
certain permutations are attainable.
Fortunately, by choosing our listing order properly, we have been
able to limit the number of attainable permutations to three. This
enables us to draw conclusions about the attainability of an arrangement
from knowledge of its associated permutation—and we can indeed then
say that only arrangements corresponding to certain permutations are
attainable.
This is the rationale behind the approach we will use to analyze
the general game. By properly choosing a listing order, we will be able
to show that only arrangements corresponding to certain permutations
are attainable and that, in fact, all arrangements corresponding to the
specified permutations actually may be attained. What is meant by
"properly choosing a listing order" will become clear shortly.

SOLITAIRE GAMES AND PUZZLES
291
Before we proceed to our analysis of the general game, let us next
consider the 2x3 version. The starting position is shown in Figure
8.35. This time, there are 360 possible arrangements, too many for us
to list them all. However, we can still associate a permutation with
each arrangement. If we adopt the listing order upper left, upper
middle, upper right, lower right, lower middle, lower left (see Figure
8.36), then the starting position gives rise to the permutation
(1,2,3,5,4).
It turns out that, of the 120 possible permutations of the numbers
1, 2, 3, 4, 5, only 60 are attainable with this listing order.
1
4
2
5
3
FIGURE 8.35
FIGURE 8.36
EVEN AND ODD PERMUTATIONS
To see why this is the case, we introduce the concept of an inversion.
In any permutation of the numbers 1, 2, ..., w, we say that an
inversion occurs each time a larger number precedes a smaller number.
Thus, for example, the permutation (1, 2, 3, 5, 4) contains one
inversion (5 precedes 4) and the permutation (1, 5, 2, 4, 3) contains
four inversions (5 precedes 2, 5 precedes 4, 5 precedes 3, and 4
precedes 3).
Definition 8.1 A permutation is said to be even if it contains an
even number of inversions, and is said to be odd otherwise.
Thus (1, 2, 3, 5, 4) is an odd permutation, (1, 5, 2, 4, 3) is an even
permutation, and (1, 2, 3, 4, 5) is an even permutation (it contains
zero inversions, and 0 is an even number).
1. [a] How many inversions are there in each of the following
permutations? Is each even or odd?
(a) (7, 1, 4, 6, 3, 2, 5) (b) (5, 1, 6, 8, 2, 4, 7, 3)
(c)(l, 2, 3, 4, 5, 6).
2. How many inversions are there in each of the following
permutations? Is each even or odd?
(a) (3, 8, 1, 4, 5, 6, 7, 2) (b) (6, 5, 4, 3, 2, 1)
(c) (3, 7, 4, 2, 8, 6, 1, 5).
PRACTICE
PROBLEMS
8.A
When we make a move, thereby changing from one permutation to
another, how is the number of inversions affected?
To answer this question, we first consider what happens when two
adjacent numbers are interchanged in a permutation—that is, if

292
CHAPTER 8
(... 5 a, 6, ...) is changed to (..., 6, a, .. .)• The position of the other
numbers appearing in the permutation relative to a and b and relative to
each other are unchanged; the only change that occurs in the number
of inversions in the permutation does so by virtue of the change in the
positions of a and b relative to each other. If a < 6, then the new
permutation (..., 6, a, ...) contains one inversion {b before a) that
the original permutation does not have; similarly, if 6 < a, then the
original permutation contains one inversion {a before b) that the new
permutation does not have. In either case, the number of inversions in
the new permutation differs from the number of inversions in the
original permutation by plus or minus 1. Thus, the two permutations
have opposite parities (one is even and one is odd).
Now suppose we want to
change the permutation
(..., a, 6, c, ...) to
(..., 6, c, a, ...), or vice versa.
This can be done by first
changing (..., a, 6, c, ...) to
(..., 6, a, c, ...) and then
changing (..., 6, a, c, ...) to
(..., 6, c, a, ...). Since each of
these changes is of the type
discussed above, each induces a
change in the parity of the
permutation. Since two such
changes occur, the new
permutation (..., 6, c, a, ...) has
the same parity as the original
permutation (..., a, 6, c, ...).
As an example, to change
from
(6,3, 1,4, 5, 2) to
(6, 1,4,3,5,2),
first change
(6,3, 1,4, 5, 2) to (6, 1,3,4,5,2)
and then change
(6, 1,3, 4, 5, 2) to (6, 1,4,3,5,2).
Note that each time we
interchange two adjacent digits,
the number of inversions
changes by H-1 or - 1.
Since two such changes occur,
(6, 1, 4, 3, 5, 2) has the same
parity as (6, 3, 1,4, 5, 2).
Generalizing, the permutation (..., a, ^j, ^2 > • • • j ^d • • •) may be
changed to the permutation (..., ^j, 62? •••> ^r> a, • • •) in r steps:
First change it to (..., b^^ a, 62 > •••> ^r> • • )> then change this to
(..., ^1, ^2 > ^) • • •) ^r 5 •. •)> and so on until (..., ^j, 62 j • • • j ^> ^r 5 • • 0 is
finally changed to (..., 61, 62 > • • > ^r> a, •. )• Since each step involves a
change of parity, (..., a, 6,, bj, ..., br, ...) and (..., b^, 62 ? • • • >
6r J ^5 • •) are of the same parity if r is even, and are of opposite parity
if r is odd.
We are now ready to see what happens when we make a move in the
2x3 game. Recall that we are using the pattern in Figure 8.36 to
determine the permutation associated with each arrangement.
Clearly, a horizontal move (when a piece slides to the left or right)
does not change the permutation at all. Neither does a vertical move
(sliding a piece up or down) in the rightmost column, as in Figure 8.37.

SOLITAIRE GAMES AND PUZZLES
293
FIGURE 8.37
A vertical move in the middle column (Figure 8.38) changes the
permutation (x, a, b, c, y) to (x, b, c, a, y) or vice versa. This is
the type of change discussed above—a is shifted two places to the
right or left. Since two is an even number, such a shift does not
change the parity of the permutation.
t
FIGURE 8.38
Similarly, a vertical move in the leftmost column (Figure 8.39)
changes the permutation (a, b, c, d, e) to (6, c, d^ e^ a) or vice versa.
This time, a is shifted four places; but, again, since four is even, no
change occurs in the parity of the permutation.
FIGURE 8.39
Thus, any move in the 2x3 version of the game does not change
the parity of the associated permutations. In other words, an even
permutation gets changed to an even permutation and an odd
permutation gets changed to an odd permutation.
Therefore, if we start with an odd permutation (1, 2, 3, 5, 4), we can
obtain only odd permutations. Since it can be shown that the number
of odd permutations is equal to the number of even permutations*,
half the possible permutations are even. Thus, half the possible
permutations cannot be obtained.
* It is not difficult to show that the number of even permutations is the same as the
number of odd permutations and that this number is therefore half the total number of
permutations. Interchanging the last two numbers of a permutation either introduces a
new inversion or else removes one, and thereby changes the parity of the permutation.
Therefore, pairing each permutation with the permutation obtained by interchanging the
last two numbers of the original permutation gives a one-to-one correspondence between
the odd permutations and the even ones:
even
(1,2,3)
(2, \ 1)
(3, 1, 2)
odd
(1,3,2)
(2, 1, 3)
(3, 2, 1)

294
CHAPTER 8
The fact that any move preserves the parity of the associated
permutations depends heavily on the way in which the permutations
are obtained—that is, on the listing order. In the 2x2 game, if we
use the listing order upper left, upper right, lower left, lower right, then
any vertical move changes the parity of the permutation, and all six
possible permutations can occur. However, if we use the listing order
upper left, upper right, lower right, lower left, then the parity of the
permutations is preserved by any move. This explains what is meant
by "properly choosing a listing order"—we choose a listing order
so that each possible move does not change the parity of the associated
permutations.
We are now ready to consider the 4x4 game.
If you haven't already solved Problem 8.4(a), page 274,
try it again now.
Solution of
Problem 8.4(a)
15
11
7
3
14
10
6
2
13
9
5
1
12
8
4
FIGURE 8.43
As in the 2x2 and 2x3 cases, we wish to consider for the
4x4 case each possible arrangement as a permutation by listing all
the numbers in some specified order. In fact, we want to choose the
listing order in such a way that any move preserves the parity of the
associated permutations. One possible listing order that works is
indicated by the path in Figure 8.40.*
With this listing order, the starting position (Figure 8.41)
corresponds to the permutation (1, 2, 3, 4, 8, 7, 6, 5, 9, 10, 11, 12,
15, 14, 13). This permutation contains 9 inversions and hence is odd.
If we can show that, using this listing order, the parity of permutations
is preserved, then only odd permutations can arise.
r
L
V
^
J
^
J
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
1
5
9
13
2
6
10
15
3
7
11
14
4|
8
12
FIGURE 8.40
FIGURE 8.41
FIGURE 8.42
Because the arrangement suggested by Loyd (Figure 8.42) gives rise
to the permutation (1, 2, 3, 4, 8, 7, 6, 5, 9, 10, 11, 12, 14, 15, 13), which
is even (8 inversions), and the arrangement of Problem 8.4(a)—see
Figure 8.43—gives rise to the permutation (15, 14, 13, 12, 8, 9, 10, 11,
* There are other patterns which give listing orders that work. This particular pattern
was chosen because the proof that it works is more easily presented than are the proofs
for most other patterns.

SOLITAIRE GAMES AND PUZZLES
295
7, 65 5, 4, 1, 2, 3), which is even (96 inversions), neither of these
arrangements is possible to achieve from the starting position indicated.
It remains to show that the suggested listing order is ''properly
chosen," that is, using this order, any move preserves parity.
As before, any horizontal move does not change the permutation
and hence certainly preserves parity.
Vertical moves may be subdivided into several cases:
First, there are those that do not change the permutation at all.
These are shown in Figure 8.44.
It
M
or
It
nH
or
Iti
i\[
FIGURE 8.44
Second, there are vertical moves which shift the number being
moved, a, two places (Figure 8.45). Thus (..., a, 6, c, ...) is changed
to (..., 6, c, a, ...) or vice versa, which does not change parity.
4
b
c
or
b
c
.ti
"li
or
a It
"M
b
c
FIGURE 8.45
The third class of vertical moves are those in which the number
being moved, a, is shifted four places (Figure 8.46). These moves
change the permutation from (..., a, b, c, d^ e, ...) to (..., 6, c, d, e,
a, ...) or vice versa, and do not change parity.
alt
%\
b
e
c
d
or
c
d
b
e
4
or
atl
%
b
e
c
d
FIGURE 8.46
Finally, the remaining vertical moves shift the number being moved
six places (Figure 8.47). Here (..., a, b, c, d, e, f, g, ...) is changed
to (..., 6, c, d, ^, /, g, a, ...) or vice versa, and again parity is
unchanged.
Thus, any vertical move preserves parity, and the argument is
complete.

296
CHAPTER 8
Liti
ni
b
g
c
f
d
e
FIGURE 8.47
d
e
c
f
b
g
4
J\
\^\i
b
g
c
f
d
e
So far, we have shown that, starting with an arrangement giving rise
to an odd (even) permutation, it is not possible to obtain one
corresponding to an even (odd) permutation; but we have not yet shown that
all arrangements corresponding to odd (even) permutations actually are
obtainable. They are! A formal proof of this fact is straightforward
but quite tedious. Rather than present such a proof, we leave it to you
to convince yourself—play with the game and try to obtain various
arrangements.
PRACTICE
PROBLEMS
8.B
1. [a] Given the arrangement in Figure 8.48, is it possible to obtain
each of the arrangements in Figure 8.49? In each case in which it is
possible, give a suitable sequence of moves.
1
2
3
4
5
4
FIG
1
3
(a)
UR]
2
5
E 8.
49
5
1
2
(b)
4
3
4
5
3
(c)
1
2
FIGURE 8.48
2. Given the arrangement in Figure 8.50, is it possible to obtain each
of the arrangements in Figure 8.51 ? In each case in which it is
possible, give a suitable sequence of moves.
FIGURE 8.50
1
5
2
6
3
7
4
1
2
3
4
5
6
7
7
3
6
2
5
1
4
4
7
3
6
2
5
1
(a)
FIGURE 8.51
(b)
(c)
We are not yet finished with Problem 8.4; we must still show how to
obtain the arrangement in Problem 8.4(b). (Note that the associated
permutation (15, 14, 13, 9, 10, 11, 12, 8, 7, 6, 5, 1, 2, 3, 4) is odd—93
inversions—and therefore the arrangement is obtainable.)

SOLITAIRE GAMES AND PUZZLES
297
Try Problem 8.4(b), page 274, if you haven't already solved it.
One way of obtaining the desired arrangement is as follows:
Sequence of moves
Start
1. 15, 14, 13, 9, 5, 1, 2, 3, 4, 8,
12 repeated five times,
followed by 15, 14, 13, 9, 5,
1, 2, 3, 4, 10
2. 11, 7, 6, 11, 10, 8
3. 11, 10, 7, 6, 10, 11, 12
Resulting position
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
(a)
12
8
10
1 "*
15
6
3
14
7
11
2
13
9
5
1
(b)
12
8
4
15
11
10
3
14
6
7
2
13
9
5
1
(c)
12
8
4
15
11
7
3
14
10
6
2
13
9
5
1
Solution of
Problem 8.4(b)
FIGURE 8.52 (d)
We do not claim that this is the most efficient procedure but we
doubt that the number of moves can be greatly decreased, if it can be
decreased at all.
1. Starting with the arrangement in Figure 8.53, obtain, if possible, PRACTICE
each of the arrangements in Figure 8.54. PPORI FK^^

298
CHAPTER 8
1
5
9
13
2
6
10
14
3
7
11
15
4
8
12
FIGURE 8.53
(a)
FIGURE 8.54
1
3
5
7
2
4
6
8
9
11
13
15
10
12
14
2
10
1
9
4
12
3
11
6
14
5
13
8
7
15
(b)
C COLORING AND THE FIFTEEN PUZZLE-
A SECOND APPROACH
Before leaving the Fifteen Puzzle, we return to a comment that we made
when we first introduced it—namely, that coloring a checkerboard has
something to do with this problem.
Suppose that the squares of the box were colored as a gray and white
checkerboard. Then the color of the square on which the blank space
falls would change with each move. Thus, if the blank space were on
a white square to begin with, then it would be on a gray square after
one move, on a white square after two moves, and so on. That is, the
blank would be a white square after an even number of moves and a
gray square after an odd number of moves.
Suppose, further, that a listing order is adopted such that each move
results in a change of parity of the associated permutations. Then,
with each move, the parity of the permutation changes, as does the color
of the square on which the blank space falls. Thus, if the starting
position is even and the blank is a white square, then, thereafter,
whenever the permutation is even, the blank will be a white square and
whenever the permutation is odd, the blank will be a gray square.
Actually, there is no nice pattern that gives such a listing order.
However, if we count the blank space as if it were the number 16,
and include 16 in each permutation, then any listing order will work.
In particular, we can adopt the listing order given by a line scanner—
the arrangement in Figure 8.55 gives the permutation (1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15, 16), which is even. We leave it to you
to verify that, with this listing order and with the blank being counted
as 16, then every move results in a change of parity.
1
9
6
14
3
11
8
FIGURE 8.55
PRACTICE
PROBLEMS
8.D
1. Verify the comment above.

SOLITAIRE GAMES AND PUZZLES
299
If the blank in Figure 8.55 starts on a white square, then an
arrangement is obtainable from the figure if and only if
(a) it gives an even permutation and the blank is on white
or
(b) it gives an odd permutation and the blank is on gray.
FIGURE 8.56
Thus,
(1,2,
Figure
3, 4, 5,
1
9
3
6 8
11
15
(a)
8.56a gives
6,7,
12,
8, 9, 10, 11
13, 15, 14,
5
16)
which is odd (1 inversion);
the blank is white, hence this is
not obtainable.
15 13
10 8
7 5
2
(b)
Figure 8.56b gives
(15, 14, 13, 12, 11, 10, 9, 8, 7,
6, 5, 4, 3, 2, 1, 16)
which is odd (105 inversions);
the blank is white, hence this is
not obtainable.
14
11 9
8 6
3 1
(c)
Figure 8.56c gives (16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1),
which is even (120 inversions); the blank is white, hence this
arrangement is obtainable.
This agrees with what we found above.
COLORED CUBES
Another puzzle in which color plays an important role is the colored
cubes puzzle of Problem 8.5. This time, however, the coloring pattern
has been imposed upon us by the statement of the problem, rather
than our being able to introduce a coloring pattern that suits our needs.
The colored cubes puzzle has been marketed by several companies
under several different titles. One marketed version was called " Instant
Insanity"—an attempt to describe the effect it might have on a
prospective solver. After all, there are 41,472 essentially different ways

300
CHAPTER 8
FIGURE 8.57
to arrange the cubes, of which only one actually yields a solution.
The order in which the cubes are stacked clearly is not important
in the solution; so we consider only arrangements with cube I on the
bottom, cube II next, and so on. There are six possible ways to choose
the front face of cube I; once it is chosen, the back face is determined
and there are four ways in which to select the top face. Hence, by the
Multiplication Principle, there are 24 different ways in which cube I
can be arranged. The same is true for each of the other cubes. Therefore,
there are 24 • 24 • 24 • 24 = 331,776 possible arrangements with cube I
on the bottom, cube II on top of cube I, etc. Because we need not
consider as different any arrangement obtained by rotating the entire
stack of four cubes (Figure 8.57), we must divide by 4, giving 82,944
arrangements. The arrangement obtained by keeping the front-back
of each cube fixed and interchanging left and right (rotating 180°
around the front-back axis) is also the same as the original. Therefore,
we must divide by 2, getting only 41,472 arrangements.
If you haven't yet solved Problem 8.5, page 275,
try it again now.
Solution of
Problem 8.5
At first, it seems as if we will have to resort to massive systematic
trial and error; no other approach presents itself. However, there are
at least two different ways of looking at the problem to treat it
mathematically.
In both approaches, the key observation is that once the position
of a particular face of the cube is determined, the position of the
opposite face is also determined. Hence, the only significant information
about each cube is which pairs of colored faces are opposite each other.
For cube I, we observe these three pairs: G-W, B-R, and R-W.
For cube II, we get: R-R, R-G, and B-W.
For cube III, we get: G-R, W-B, and G-W.
And for cube IV, we get: G-G, B-R, and B-W.
We are now ready to present the first of the two solutions.
Suppose we assign numbers to each color as follows: 1 to blue,
2 to red, 3 to green, and 5 to white.
We can then assign to each pair of opposite faces the product of
the numbers assigned to the faces themselves. We thus obtain the chart
in Figure 8.58.
When the cubes are stacked properly, there will be a B = 1, R = 2,
G = 3, and W = 5 on each side of the stack, giving a product of
l-2-3-5 = 30.In particular, the front and back of the column must
each give products of 30, yielding a total product of 30 • 30 = 900 for
the front and back faces. But the front and back of the column are
made up of pairs of opposite faces from each cube. Hence, we must

SOLITAIRE GAMES AND PUZZLES
301
opposite face
cube I
cube II
cube III
cube IV
G-W
B-R
R-W
R-R
R-G
B-W
G-R
W-B
G-W
G-G
B-R
B-W
FIGURE 8.58
corresponding product
3-5 = 15
1-2 = 2
2-5 = 10
2-2 = 4
2-3 = 6
1-5 = 5
3-2 = 6
51 = 5
3-5 = 15
3-3 = 9
1-2 = 2
1-5 = 5
choose one pair of opposite faces from each cube so that the product
of the products corresponding to the chosen pairs is 900. The same
is also true of the side faces of the stack.
Hence, the problem reduces to selecting one pair of opposite faces
from each cube so that the associated product is equal to 900 (to
represent front and back of the stack) and then selecting a different
pair of opposite faces from each cube so that the associated product
is again 900 (representing the sides of the stack).
We first consider all possible products of one pair of opposite faces
from cube I and one pair from cube II. We obtain Figure 8.59.
We could now multiply each of these by each of the possible products
for cube III and then multiply by the possibilities for cube IV, to see
which products give 900. It is a little easier first to compute the
possible products for cubes III and IV alone, and then to see which of
these multiplied by which of the ones above yields 900. We get the
table in Figure 8.60.
The cases in which the necessary multiplier is found in the list for
cube I
15
15
15
2
2
2
10
10
10
cube II
4
6
5
4
6
5
4
6
5
FIGURE
= 60
= 90
= 75
= 8
= 12
= 10
= 40
= 60
= 50
8.59
cube III
6
6
6
5
5
5
15
15
15
cube IV
9= 54
2 = 12
5 = 30
9 = 45
2= 10
5 = 25
9= 135
2 = 30
5 = 75
multiplier needed to give 900
impossible
75 >/
30 (not in the I-II list)
20 (not in the I-II list)
90 >/
36 (not in the I-II list)
impossible
30 (not in the I-II list)
12 V
FIGURE 8.60

302
CHAPTER 8
cubes I and II have been checked, indicating that there are three ways
in which a pair of opposite faces from each cube can be chosen so
that the product is 900. These are shown in Figure 8.61.
combination 1
combination 2
combination 3
cube I
15(G-W)
15(G-W^
2(B-R)
cube II
5(B-W)
6(R-G^
6(R-G)
cube III
6(G-R,
5(W- B)
15(G-W)
cube IV
2(B-R)
2(B-R)
5(B-W^
FIGURE 8.61
We must select one of these combinations to represent the front-back
of the stack and another combination to simultaneously represent the
sides. But combinations 1 and 2 cannot be used simultaneously (they
both make use of the G-W pairing on cube I) and combinations 2
and 3 cannot be used simultaneously (they both use the R-G pairing
on cube II). Hence, we must use combinations 1 and 3.
Since we are considering rotation of the stack to be a symmetry—
that is, not to yield a new solution—there is no loss of generality in
assuming that combination 1 represents front-back and that the green
face on cube I is in front. This forces the white face of cube II, the red
face of cube III, and the blue face of cube IV to be in front.
Combination 3 now shows us how to arrange the side faces of the
cubes, without affecting front and back. There are two solutions,
depending on whether the blue face of cube I is on the right or left.
(Actually, the solutions are equivalent, because one can be obtained
from the other by turning each cube upside down without changing
front or back.) The two solutions we get are shown in Figure 8.62.
^
B— B W^G
w
G— ^ y
R— B
W
G
R
B
y^
w
V
/
w
cube I
cube II
cube III
cube IV
front
G
W
R
B
back
W
B
G
R
solution 1
right side left side
B
R
G
W
R
G
W
B
solution 2
right side left side
R
G
W
B
B
R
G
W
FIGURE 8.62

SOLITAIRE GAMES AND PUZZLES
303
Before we leave this approach to the solution of Problem 8.5, we
should comment on the reason why we chose the numbers 1, 2, 3,
and 5 to represent the various colors. We want the numbers to be
relatively small so that they are easy to work with; and we want them
to be pairwise relatively prime (no two have a common factor other
than 1) so that each possible color combination corresponds to a
different product. For example, suppose we used 1, 2, 3, and 4 instead
of 1, 2, 3, and 5. And suppose we found that we obtain the proper
product (which in this case would be 576) when cube II has value 4
and when the other three cubes are properly chosen. We would now
have to consider two possible arrangements of cube II, because we
could get 4 in two ways—B-W (1 • 4) or R-R (2 • 2). By choosing
numbers that are pairwise relatively prime, we help limit the
possibilities.
^ COLORED CUBES-A SECOND APPROACH
We now turn our attention to a second approach to the solution of
Problem 8.5. This time the approach is graph-theoretical.
We associate a graph with each cube as follows: We first represent
the four colors—blue, red, green, and white—as points (or nodes). We
then draw lines (edges) connecting points that correspond to opposite
faces of each cube in Figure 8.4. We obtain Figure 8.63.
cube I cube II cube III cube IV
R<
FI(
I (
1 1
3URE 8.63
G
W
•S^
B»-
-—.G
—-W
R«
• G
B«
•W
Rt
B^
^W
We now superimpose these graphs on one diagram (Figure 8.64).
FIGURE 8.64

304
CHAPTER 8
When the cubes are stacked properly, each color appears once in
front and once in back. We therefore can select one opposite pair from
each cube (one edge from the graph of each cube) such that each
color (each node) is used exactly twice. In other words, we can select
four lines (one corresponding to each cube) from the composite graph
and assign an orientation to each (designate one end or direction to
represent the front and the other end or direction to represent the back)
so that each of the four points has one selected line entering it and one
leaving it. It is not difficult to see from the composite diagram that
this can be done only as shown in Figure 8.65.
Ill
R
IV
t >A
B II W
1 2 3
FIGURE 8.65
Again we must select two of these—one for the front-back of the
stack and one for the sides. But, as before, combination 2 cannot be
used with 1 (they both use the R-B pair from cube IV) or with 3
(they both use the R-G pair from cube II). Hence, we must use
combinations 1 and 3.
Putting G in the front of cube I and B on the right, we obtain the
situation shown in Figure 8.66.
cube I
cube II
cube III
cube IV
front
G
W
R
B
FIGURE 8.66
back
W
B
G
R
right side
B
R
G
W
left side
R
G
W
B
This is the same as solution 1 that we found above.
Similarly, putting G in the front of cube I and B on the left, we
obtain the other solution found above.
THE CHAPTER IN RETROSPECT
This chapter has presented some solitaire games that have mathematical
solutions, and it has shown that mathematics pops up in unusual places.
In solving the problems of this chapter we have made use of

SOLITAIRE GAMES AND PUZZLES
305
mathematical induction, geometry, coloring arguments and results from
number theory, graph theory, and permutation theory.
In the case of some of the puzzles, you might ask why a particular
approach is used. The answer in each case is that it works. How the
original solvers of these problems were led to these ideas is sometimes
a mystery. Often, insight and discovery come as a result of
experimentation, or just playing around. There are no known bounds to
human ingenuity.
Now try to apply what you have learned to the exercises that
follow. And, no matter what your field of endeavor, be on the lookout
to see if mathematics can be of help to you.
Exercises
8.1. In the Tower of Brahma puzzle, label
the three posts A, B, and C. If we wish to
transfer n rings from post A to post C in as
few moves as possible, to which post should we
make the first move ? (Note: The answer will
depend on n.)
(c) There are five switches in the row?
(d) There are six switches in the row?
-¥ -¥ (e) There are n switches in the row
(w odd)?
-¥-¥ {T) There are n switches in the row
(w even)? (See [29].)
8.2. [h] [a] In order to prevent tampering by
unauthorized individuals, a row of switches at
a defense installation is wired so that, unless
the following rules are followed in
manipulating the switches, an alarm will be activated:
1. The switch on the right may be turned
on or off at will.
2. Any other switch may be turned on or
off only if the switch to its immediate right
is on and all other switches to its right are
off.
What is the smallest number of moves in
which such a row of switches, which are all on,
may be turned off without activating the alarm
if:
(a) There are three switches in the row?
(b) There are four switches in the row?
-¥ 8.3. [a] Another row of switches at the
defense installation is governed by the
following rules of movement:
1. Any switch may be turned on at will;

306
CHAPTER 8
however, doing so activates a relay that
automatically turns off the switch to its
immediate left (unless that switch is already
off).
2. The switch on the right may be turned
off" at will; any other switch may be turned
off only if the switch to its immediate right
is on and all other switches to its right are
off.
What is the smallest number of moves in
which such a row of switches, which are all on,
may be turned off if:
(a) There are three switches in the row?
(b) There are four switches in the row?
(c) There are five switches in the row?
(d) There are six switches in the row?
(e) Etc., up to ten?
(c) The area of any triangle (Figure 8.70)
is |6/j, where b is the length of one side and
h is the length of the altitude to that side.
\bh
FIGURE 8.70
"^ (d) [a] The area of a trapezoid (Figure
8.71) is |/i(6i + ^2) where b^ and 62 are the
lengths of the parallel sides, and h is the length
of the perpendicular distance between them.
r/z (6, + 62)
FIGURE 8.71
Dissection Problems
8.4. Given that the area of a rectangle
(Figure 8.67) with base of length b and
A == bh
FIGURE 8.67
height of length h is bh, use a dissection
argument to show that:
(a) [a] The area of a parallelogram (Figure
8.68) with base b and altitude h is bh.
Nl/
A ^ bh
FIGURE 8.68
(b) The area of a right triangle (Figure 8.69)
with legs of length 6 and h is ^bh.
A = -^bh
FIGURE 8.69
8.5. [h] (a) Given Figure 8.72, make two cuts
and use the pieces to form a square.
FIGURE 8.72
(b) [a] Given Figure 8.73, cut it into the
fewest number of pieces that can be
reassembled to form a right triangle.
FIGURE 8.73
(c) Give proofs of the Pythagorean Theorem
suggested by each of the diagrams in Figure
8.74. ([14], pp. 72-73)
8.6. [h] [a] Given a
(a) 2 X 2 checkerboard,
(b) 4 X 4 checkerboard,
(c) 3 X 3 checkerboard with central cell
missing.

SOLITAIRE GAMES AND PUZZLES
307
(i)
FIGURE 8.74
(ii)
(d) 5 X 5 checkerboard with central cell
missing,
(e) 3 X 3 checkerboard with corner cell
missing,
in how many ways can each of the above
boards be cut along the lines of the
checkerboard into two identical pieces if:
(i) coloring is counted and pieces cannot
be turned over;
(ii) coloring is not counted but pieces
cannot be turned over;
(iii) coloring is not counted and pieces may
be turned over?
Whether or not two pieces are considered
to be identical depends on whether (i), (ii),
or (iii) applies. Similarly, two ways of cutting
the board will be considered to be the same if
the pieces obtained from cutting in one way
are congruent to the pieces obtained by cutting
in the other way subject to the conditions in
(i), (ii), or (iii), as the case may be.
8.7. Given a
(a) 2 X 2 checkerboard,
(b) 4 X 4 checkerboard,
(c) 3 X 3 checkerboard with central cell
missing,
•It (d) 5 X 5 checkerboard with central cell
missing,
(e) 3 X 3 checkerboard with corner cell
missing.
in how many ways can each of the above boards
be cut along the lines of the checkerboard into
4 identical pieces if:
(i) [a] coloring is counted and pieces
cannot be turned over;
(ii) coloring is not counted but pieces
cannot be turned over;
(iii) [a] coloring is not counted and pieces
may be turned over?
Two cuttings are considered to be the same
as described in Exercise 8.6.
Tangrams
8.8. [a] Seven pieces make up the tangram
square (Figure 8.75).
Trace this figure,
cut out the pieces
and use them for
Exercise 8.8 and
for other tangrams.
FIGURE 8.75
(a) Fit the above seven pieces to form the
letter T in Figure 8.76. ([53], Problem 20)
(b) Fit the above seven pieces to form the
boat in Figure 8.77. ([53], Problem 169)

308
CHAPTER 8
FIGURE 8.76
FIGURE 8.77
Polyominoes (S. Golomb)
8.9. (a) [h] [a] Can an 8 x 8 checkerboard
with one red and one black cell missing be
covered by 31 dominoes, regardless of which
red and black cells are missing? Justify your
answer.
(b) [h] [a] Can an m x n checkerboard
(with m and n odd) that has one cell removed
be covered with dominoes?
(c) [h] [a] Can the 5 tetrominoes (Figure
8.12, page 282) be used to form a rectangle?
(d) [s] Draw the 12 pentominoes (5-cell
shapes).
-¥ (e) [a] Find the smallest region on the
8x8 checkerboard into which each of the 12
pentominoes, taken one at a time, will fit.
^ ♦ (f) [h] [a] Arrange the 12
pentominoes to form two 5x6 rectangles.
(g) [h] Show that the region in Figure 8.78
cannot be covered by the 12 pentominoes.
FIGURE 8.78
(h) [a] How many hexominoes (6 cells) are
there?
(See [26], pp. 30, 33, 37)
Polycubes
8.10. (a) [a] Consider a 3x3x3 cube
made up of 27 unit cubes. Remove one of these
unit cubes and cover the rest of the
configuration with 2x1x1 dicubes. Which unit cube
could be removed in order to accomplish this?
(b) Can the 3x3x3 cube be made up of
9 of the tricubes shown in Figure 8.18
page 285)?
(c) Show by a coloring argument that a
2x3x4 solid cannot be made from the
polycubes numbered (1), (2), (3), (4), (5), and (7)
in Figure 8.17 (page 285).
Peg Solitaire
8.11. (a) [3 [a] In Hi Q (Problem 8.3, page
274), which of the original pegs could possibly
be the one remaining in the center?

SOLITAIRE GAMES AND PUZZLES
309
(b) [h] In the French version of Hi Q (see
Figure 8.79), show that you cannot start with
the center hole empty and end with one peg
anywhere.
checkers, and the jumped peg is removed. Can
you leave only one peg?
FIGURE 8.79
(c) [h] In the French version, show that no
matter which hole starts empty, you cannot
end with one peg in the center hole.
(d) [h] [a] For which holes other than the
center do the statements in (b) and (c) hold
true?
(e) [a] On the regular board, start with
Figure 8.80 (the cell numbering is as in the text
of this chapter, page 287) and get one in the
center.
24
33 34 35
44
54
FIGURE 8.80
(f) Start with Figure 8.81 and get one in the
center.
24
34
42 43 44
54
64
FIGURE
(g) Start with Figure
the center.
34
43 44
52 53 54
FIGURE
45 46
8.81
8.82 I
45
55 56
8.82
• • • •
• • • •
• • • •
• • • •
FIGURE 8.83
(i) [h] Given the board in Figure 8.84,
pegs (indicated by •) jump toward or away from
the center or along the circle the peg is on. The
jumped peg is removed. A piece cannot jump
into the center except on the last move. Show
that you cannot end with one peg in the center.
(h) [h] Given the board in Figure 8.83,
pegs (indicated by •) jump diagonally as in
FIGURE 8.84
(j) [a] On a 5 X 5 board with the 9 center
squares filled, you may jump horizontally,
vertically, or diagonally (removing the jumped
peg). Successive jumps by the same peg count
as one move. What is the fewest number of
moves that allows you to leave one peg in the
center?
(k) [h] [a] Can (j) be done using only
horizontal and vertical jumps?

310
CHAPTER 8
(1) [h] The board is a 10-cell triangle (see
Figure 8.85). Move as in Hi Q, with horizontal
and diagonal jumps. Show that, if you begin
with any corner cell empty and the remaining
cells occupied, it is not possible to leave only
one peg remaining.
FIGURE 8.85
(m) [h] [a] The board is a 15-cell triangle
(see Figure 8.86). Leave the top cell vacant.
Move as in Hi Q, with horizontal and diagonal
jumps, so that only one peg remains. Which
cell could contain the last peg?
FIGURE 8.86
8.12. [a] Consider the arrangement of black
and white pegs in a row shown in Figure
8.87a, with the center cell vacant. White pegs
move to the right; black, to the left. A peg may
advance one space to a vacant hole, or may
jump over another peg to an adjacent vacant
hole. The jumped peg is not removed.
^^^ •••
(a)
• •• ooo
(a) What is the fewest number of moves
required to reverse the black and white pegs;
that is, to attain the position in Figure 8.87b?
(b) The same as (a) but with n black and
n white pegs.
(c) Given the arrangement of black and
white pegs shown in Figure 8.88, white pegs
move to the left or up, black pegs move to the
right or down. Otherwise the rules are the same
as in (a). How many moves are required for the
black and white pegs to interchange positions?
•
•
•
•
•
•
•
•
•
•
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o '
o
o
o
o
o
o
FIGURE 8.88
M M (d) The same rules as (a), but with
8 black and 8 white pegs arranged on the
board in Figure 8.89, the black pegs moving
down and to the right and the white moving
up and to the left.
• • •
• • •
^ n^ n^
ooo
(b)
FIGURE 8.87
FIGURE 8.89
8.13. Given 2n checkers in a straight line,
the object of this puzzle is to obtain n stacks
of two checkers each by making n moves of
the following type: One checker may be moved
(to the right or left) over two other checkers
so that it lands on top of the next checker in
line. (Note that stacks count as two checkers
and empty spaces do not count as any
checkers.)

SOLITAIRE GAMES AND PUZZLES
311
Explain how this may be accomplished for
each of the following values of n.
(a) w = 4
(b) [1 n = ^
(c) [h] General n.
Sliding Block Puzzles
8-14. [h] Some variations of the Fifteen
Puzzle use letters instead of numbers.
Consider the 3x3 box shown in Figure
8.90, with the center empty and with the word
ROTATION spelled clockwise around the
perimeter.
R
N
O
O
I
T
A
T
FIGURE 8.90
Move the pieces as in the Fifteen Puzzle
so that the word ROTATION is again
spelled in a clockwise direction around the
perimeter, but the letter R is in the middle
box of the top row.
8.15. [a] When Foster Quarrels died, he left
a will that provided that his estate should be
divided equally among his five cousins, Ann,
Dan, Jan, Nan, and Van. Unfortunately, the
cousins did not get along well with each
other. In fact, they detested each other so
vehemently that it would have been unwise to
allow any two of them to be in the same room.
It came time to read the will at the
apartment of the late Mr. Quarrels, which is
pictured in Figure 8.91. The lawyer placed
Ann in the kitchen, Dan in the dining room.
dining
room 1
1
kitchen
1 1
FIGURE 8.91
den
foyer
1 1
bedroom
1
1
bathroom
1 1
Jan in the den. Nan in the bedroom, and
Van in the bathroom. However, Van insisted
that he wait in the kitchen, and Ann wanted
to go to the bathroom. How could the lawyer
make this change in the fewest possible moves,
without any two cousins meeting each other
in the process? (The other cousins need not
end up in the same rooms in which they
were originally placed.)
Colored Cubes
8.16. (a) Suppose that we keep the first three
cubes in Problem 8.5 (Figure 8.4, page 275),
but that cube IV is replaced by the cube in
Figure 8.92a. Show that it is not possible to
stack the cubes as required in Problem 8.5.
w—-
FIGUR
B
t
W
/
W
(a)
E 8.92
/
w-
r
\
G
(b)
(b) Suppose that we keep the first three
cubes in Problem 8.5, but that cube IV is
replaced by the cube in Figure 8.92b. Find all
possible ways of stacking the cubes as required
in Problem 8.5.
8.17. [h] [a] (P. A. MacMahon) Given 6
colors, there are 30 different ways to color a
cube so no two faces of the cube receive the
same color.
Select any one of the 30 cubes—say^ the

312
CHAPTER 8
cube that is blue on top, yellow on the bottom,
red in front, white in back, green on the left,
and orange on the right. Use 8 of the remaining
29 possible cubes to form a 2 x 2 x 2 model
of the chosen cube. In constructing such a
model, any faces of the unit cubes that touch
each other must be of the same color.
Other Solitaires
8.18. [h] Use a coloring argument to show
that a reentrant knight's tour (see Chapter 6)
is not possible on a board having an odd
number of cells.
8.19. [a] Several modern games (Oware or
Kalah, and Ruma are examples) are variants
of the game Mancala, whose origin can be
traced back more than 3000 years. The
following is an interesting solitaire version of this
game.
One bowl and n cups are placed in a circle,
and k marbles are placed in each cup (but none
are put in the bowl). Taking the marbles from
any cup, "sow" them, one at a time, in a
clockwise direction around the circle, starting
with the cup (or bowl) after the one from which
the marbles were taken and ending when the
marbles are all used up. Note that, if the
number of marbles is sufficiently large, you
may proceed around the circle more than once,
the bowl and original cup receiving one (or
more) marble(s).
If the last marble lands in an empty cup,
you lose. If the last marble lands in a nonempty
cup, you must use the marbles from that cup
for sowing next. If the last marble lands in the
bowl, you may select any cup to sow from.
Following these rules, the object of the game
is to transfer all the marbles to the bowl.
Can this be accomplished if:
(a) w = 3, /^ = 1? {b)n = \k = 2'>
(c)n = 3, k = 3?
(c) n = 4, k = 2?
(g)n = 4,k = 4?
(i) n = 6, k = 2?
(d)n = 4, k= 1 ?
(f) n = 4, k = 3?
(h)n = 5, k = 2?
(Adapted from [10], Problem 185)

Potpourri
In previous chapters, we considered a wide variety of recreational
problems, with each chapter devoted to problems relating to a particular
area of mathematics. But we have only scratched the surface of the many
topics usually considered to be part of recreational mathematics. It
is not possible to include all of this material in any one book; but
we would like to present here, in our final chapter, a sampling of some
well-known recreational problems that do not require any specific
mathematical tools. All that is needed to solve them is some careful
thought.
There is no text for this chapter, other than the brief remarks
that introduce some of the problems. The chapter is what its name
says—a potpourri.
Decimation
The term " decimation " stems from an ancient
custom of executing one-tenth of a mutinous
crew by lining them up and selecting every
tenth man for the gallows. However, the
prototype of this kind of problem, often
referred to as the Josephus Problem, is based
on an event that occurred in the year 67.
After the fall of Jotapata, during the rebellion
against Rome, Josephus and forty other Jews
fled and took refuge in a cave. It was decided
that they would kill themselves rather than
allow themselves to fall into Roman hands.
Josephus, who was not ready to die, devised
the following plan. He arranged the forty-one
people (including himself) in a circle and
suggested that, counting clockwise, every third
person should be killed, until only one person
remained. That person would then commit
suicide. Josephus placed himself and a friend
so that they would be the last two survivors.
9.1. In what positions relative to the one from
which the counting started did Josephus place
himself and his friend?
313

314
CHAPTER 9
9.2. Nine children are seated in a circle
playing the game of Buzz. The game starts with
one player saying "one"; the person on his or
her right says "two"; the individual on that
person's right says "three"; and so on,
continuing around and around the circle.
Whenever anyone is about to say a number
that contains a seven as one of its digits or
that is divisible by seven, the person says
"buzz" instead of the number. Failure to do
so eliminates that person from the game and
removes him or her from the circle. The next
player then resumes the count with the next
number.
The children have no trouble recognizing
numbers that contain the digit seven, but they
do not yet know the multiples of seven.
Assume that everyone who comes to a multiple
of seven that does not contain the digit seven
is eliminated.
Where is the eventual winner of the game
seated with reference to the player who starts
by saying "one"?
9.3. [a] The magician removed ten cards,
with face values ace through ten, from a deck
of playing cards. He arranged them in some
order and held them in a pile in his hand,
face down. He then spelled aloud the word
"ace"—A C E—removing one card from the
top of the deck and placing it on the bottom
(often referred to as "ducking a card") for
each letter. He then turned the next card face
up and placed it on the table. It was the ace.
Using the nine cards remaining in his hand.
he spelled TWO, again ducking one card
with each letter. The next card was then placed
face up on the table. It was the two. He
continued in this manner, ducking a card with
each letter in the face value of the next card
to be exposed, until only the ten remained in
his hand.
How had he originally arranged the cards?
9.4. At Helen's birthday party, she and
eleven friends sat in a circle playing Hot
Potato. The rules of the game call for passing
a potato in a clockwise direction while music
plays. When the music stops, the person
holding the potato is eliminated and must leave
the circle. Unfortunately, the record player at
Helen's house was broken, so Helen's mother
was asked to sing. Not wishing to do so, she
mentioned a number, w, greater than ten, and
suggested that every wth player to handle the
potato should be out. The counting started
with Helen and moved clockwise.
After six players had been eliminated,
Jeffrey realized that he was going to be the
next to go. Quick as a wink, he suggested that
passing in a clockwise direction is unfair to
lefties and they should therefore pass
counterclockwise. Everyone agreed, and Jeffrey
eventually was the winner.
What was the smallest number, w, greater
than ten, that Helen's mother could have
suggested; and where was Jeffrey sitting in
relation to Helen?
9.5. Ten teenagers got together to play
basketball. Lauren and Lisa were appointed
team captains and had to choose their teams.
The eight remaining players happened to be
standing in a circle.
"How should we choose?" asked Lauren.
Lisa suggested the following scheme:
"Starting with Burt as number one, I'll take
the fourth person, counting around the circle
in a clockwise direction. You'll take the fourth
person after that. I'll take the fourth after

POTPOURRI
315
that, and so on. Each person will leave the
circle as soon as he or she is selected."
"That isn't fair," objected Lauren. "You
get all the good players that way. Let me get
the first player that's counted out, you get the
next player, and so on."
"OK," replied Lisa. "But let me pick a
number other than four."
Lauren agreed.
What number should Lisa pick so that she
still gets the good players?
Coin Weighing
9.6. [a] Clu D. Tector was lecturing on
methods for detecting counterfeit coins. To
illustrate a point, he removed nine coins from
his pocket and said that, although the coins
appeared to be identical, one was counterfeit
and lighter than the rest. He produced a
balance scale and challenged the audience to
suggest a plan for determining, in the fewest
possible weighings, which of the coins was the
counterfeit.
When no one found the optimal scheme,
Clu mentioned that two weighings should
suffice.
Can you figure out how to do it?
9.7. [s] After Clu (see Exercise 9.6)
demonstrated how two weighings would suffice, one
of the members of the audience posed a similar
problem: Removing four coins from her purse,
she said that one of them was counterfeit and
did not weigh the same as the other coins, but
she would not say whether it was heavier or
lighter. She then produced a fifth coin and
said it was a good coin. The problem she
posed was to devise a plan for finding the
counterfeit coin in no more than two
weighings.
Clu did it. Can you?
'^^ 9.8. [h] Referring to the problem posed in
Exercise 9.7, Clu said, "Let's consider a more
difficult variation. If you have twelve coins,
one of which is either heavier or lighter than
the rest, can you find the counterfeit coin in
no more than three weighings and determine
whether it is heavier or lighter?"
M M 9.9. [h] Clu placed five coins on the
table. They appeared to be identical, but Clu
remarked that they were all made of different
alloys and that no two weighed the same.
In no more than seven weighings on a
balance scale, can you determine which coin
is lightest, which is next lightest, and so on?
M M 9.10. \h\ For his next weighing
demonstration, Clu told his audience that the five
coins (see Exercise 9.9) weighed 10 grams,
20 grams, 30 grams, 40 grams, and 50 grams.
He then proceeded to demonstrate how one
can determine, in no more than five weighings
on a balance scale, which coin is which.
How did he do it?

316
CHAPTER 9
9.11. As his final topic for the evening, Clu
pointed out that a regular scale, as distinct
from a balance scale, can often be even more
helpful. He displayed five sacks of coins and
said that four of the sacks contained ordinary
coins, weighing 1 ounce each, but that the
remaining sack contained counterfeit coins,
each weighing 2 ounces.
The problem Clu posed is to determine in
one weighing which is the sack of counterfeit
coins.
How can this be done if:
(a) [a] The scale you use shows the weight
correct to the nearest ounce?
(b) [h] The scale indicates only pounds?
Shunting
In Chapter 1, we considered some shunting
problems in which the object was to switch
the position of a locomotive and railroad cars
on a track. We now present several other
variations of this type of problem.
9.12. An accident on the main track between
Eastville and Westville forced the Eastville-
Westville local to take a seldom used side track.
Unfortunately, three freight cars had been left
standing on this track. Fortunately, there was
a triangular arrangement of tracks in the
vicinity (Figure 9.1), but only one car or
engine could fit on any of the four parts
(1, 2, 3, 4) of the arrangement at one time.
The engine D of the local can go in either
direction, pulling or pushing cars in the
process. Every time the engine reverses its
direction of movement, fuel is wasted; hence,
the engineer wants to pass the stalled cars
with as few engine reversals as possible.
(a) If the engineer, Connie Duktor, wishes
to leave the three freight cars exactly as she
found them, and she also wants to arrive at
Westville with her train exactly as it was when
she started (with cars in the original order and
facing the original direction), how can she pass
the freight cars with no more than 22 engine
reversals?
(b) [s\ If she does not care in what order
or direction she leaves the stalled cars on the
main track but does want to arrive in
Westville with her train exactly as it was when she
left Eastville, how can she pass the freight
cars with no more than 16 engine reversals?
9.13. [a] After passing the freight cars (see
Exercise 9.12), the Eastville-Westville local
continued on its way. Unfortunately, the
Westville-Eastville local was using the same
track. Fortunately, there was a siding (Figure
9.2). Unfortunately, the siding had room
enough for only one car or engine at a time.
How can the trains pass with the fewest
total engine reversals?
j[ 0~'_^ P"! 0~ "_: 9"! 0~ ".^ yf-^'"^
^a^Q'io"-'i^Lo-'- yc
ABC
FIGURE 9.1
D E

POTPOURRI
317
3>o~"— 9\p~."— yfo --IJC
iSzioito^TI?! o~ - ';^C
A B
FIGURE 9.2
I^St
tunnel
FIGURE 9.3
9.14. [h] On the track in Figure 9.3,
interchange the position of cars A and B and
return the engine to its starting position. By
the way, only the engine can pass through the
tunnel; the railcars are too big.
Syllogisms
9.15. All bangs are bengs.
All bengs are bings.
All bings are bongs.
Can we conclude that all bangs are bongs?
9.16. [s] All gangs are gengs.
Some gengs are gings.
No gengs are gongs.
Can we conclude that
(a) No gangs are gongs?
(b) No gings are gongs?
(c) Some gings are gongs?
(d) Some gangs are gings?
9.17. All Cyclopses have only one eye.
People cannot get a driver's license unless
they have depth perception.
No one eyed creatures have depth
perception.
Polyphemus is a Cyclops.
zsEil
engine
Is it possible that Polyphemus can get a
driver's license?
9.18. [a] According to Lewis Carroll in The
Hunting of the Snark (1876):
All Boojums are snarks.
Every Bandersnatch is a frumious animal.
Only animals which frequently breakfast at
five o'clock tea can be snarks.
No frumious animals breakfast at five
o'clock tea.
Are any Bandersnatches Boojums?
9.19. [s] [a] It is true that:
1. All colonial hymenopters are capable
of reproducing by parthenogenesis.
2. All bees live in hives.
3. Every insect that helps pollinate fruit
orchards is beneficial to the farmer.

318
CHAPTER 9
4. Only colonial hymenopters live in
hives.
5. All bees help pollinate fruit orchards.
Can we conclude that all insects that are
beneficial to the farmer are capable of
reproducing by parthenogenesis?
9.20. [a] Nobody who would walk in a
cemetery after midnight is superstitious.
All people who believe that knowledge is
obtainable through reason are rationalists.
Everyone who walks under a ladder is taking
a chance that a bucket of paint might fall on
his or her head.
Only people who believe in the supernatural
are afraid of ghosts.
Only people who are superstitious do not
walk under ladders.
No one who is a rationalist believes in the
supernatural.
Only people who are afraid of ghosts would
not walk in a cemetery after midnight.
Is there any relationship between people
who would take a chance that a bucket of
paint might fall on their heads and people
who believe that knowledge is obtainable
through reason?
9.21. Everyone who believes that mermaids
exist has visited Atlantis.
Only people who have found favor with the
gods are protected by Neptune.
No one who has not tasted real ambrosia
has partaken of food with the gods.
People who do not believe that mermaids
exist have never seen a mermaid.
Everyone who has tasted real ambrosia has
a hangover.
Only people who have partaken of food
with the gods have ever been invited to Mount
Olympus.
No one who has not survived a shipwreck
has visited Atlantis.
Everyone who has found favor with the gods
has been invited to Mount Olympus.
No one who is not protected by Neptune
has survived a shipwreck.
What, if anything, can we conclude about
the connection between seeing a mermaid and
having a hangover?
Grab Bag
9.22. [a] a jigsaw puzzle contains 500 pieces.
A "section" of the puzzle is a set of one or
more pieces that have been connected to each
other. (Note that a single piece by itself is
also called a section.) A "move" consists of
connecting two sections.
What is the smallest number of moves in
which the puzzle can be completed? ([41],
Vol. 26, p. 169)
9.23. A bacterium in a petri dish filled with
agar-agar will divide in two every 20 minutes,
forming two bacteria. Starting with one
bacterium, it takes 24 hours to completely fill
the petri dish with bacteria.
How long does it take until the petri dish
is half full?
9.24. [h] [a] Fi Bonacci breeds a strain of
self-fertilizing worms that reproduce in the
following manner. Each worm produces its
first offspring at the age of two weeks and
continues to produce exactly one offspring
each week thereafter.
If Mr. Bonacci started breeding these worms
just over twelve weeks ago, and if he had
exactly one newborn worm at that time, how
many worms does Fi have now, assuming that
none has died?

POTPOURRI
319
9.25. [a] a phonograph record of radius 15
centimeters is playing on a turntable. The
grooves on the record are I millimeter wide.
The grooves begin 5 millimeters from the edge
of the record and end 5 centimeters from the
center.
How far does the needle travel during the
time that the record is playing?
9.26. [a] a tired old inchworm fell into a
gopher hole that is 10 inches deep. The
worm started climbing up the side of the hole,
progressing at the rate of 1 inch every 5
minutes. At the end of each 15 minutes of
climbing, the worm had to pause to rest for
1 minute. During each of these rest periods,
the worm slipped back 1 inch.
How many minutes does it take the worm
to climb out of the hole?
9.27. [a] a man puts all his socks in a drawer,
without pairing them. He owns two kinds of
socks, black ones and brown ones. All the
black ones match each other, and all the brown
ones match each other. One day, the man is
in a hurry and, without looking to see what
color they are, he pulls some socks from the
drawer. If the drawer contains 8 black socks
and 12 brown ones:
(a) What is the smallest number of socks
he must pull in order to be sure that he has
a pair of black socks ?
(b) What is the smallest number of socks
he must pull in order to be sure that he has
at least one pair of socks of either color ?
9.28. Paula Peedle's poodle Puddles is the
best dressed dog in town. Not only does he
have fancy outfits for all occasions, but he
even has knitted socks to match. In all. Puddles
has 12 red socks, 24 green ones, and 40 white
ones, which Paula keeps in a large laundry bag.
Whenever Paula dresses Puddles, she first
draws the socks, one at a time, from the bag,
until she has 4 socks of the same color. She
then chooses the rest of Puddles' outfit to
match the socks.
What is the greatest number of socks Paula
ever has to draw from the bag?
9.29. [a] Hardluck Harry Harris was shooting
craps with Shifty Steve Sutter. Steve had the
dice. On his first pass, he rolled a 7 (see
Figure 9.4a). On his next pass, Steve threw an
11. (Figure 9.4b).
(a)
TT7|»
(b)
FIGURE 9.4

320
CHAPTER 9
"You cheat," shouted Harry. "Give me
back my money."
Harry had no reason to think that all dice
are made the same way.
How then did he know that Steve was
cheating ?
9.30. [a] Find the largest set of integers less
than 100 such that it is not possible to choose
a subset that sums to 100. ([15], Problem 36)
9.31. The ends of a piece of string are to
be tied together so that the string will form
a circle with circumference of 12 inches. Before
the ends are tied, three knots are to be placed
in the string at such locations that, after the
string is tied (thereby creating a fourth knot),
every integral number of inches is the distance
along the circumference of the circle between
two of the knots.
(a) [a] Where should the three knots be
tied?
(b) What if the circumference is to be 19
inches and we can make four knots before the
ends are tied?
THE BOOK IN RETROSPECT
In this book we have presented some of the wide variety of problems
that are usually dealt with in books on mathematical recreations. We
have also presented some related mathematical topics, to illustrate
how mathematics can be applied in solving problems and also how
mathematical theory can be motivated and developed. In the course
of working through this book, your abilities in general problem solving
have been exercised and therefore strengthened and you will be able
to carry them over to problem solving in other areas.
We hope that you will return to this book from time to time, to
try problems you have not yet attempted or to find better solutions
for those that you have already solved.
Although this book contains over 450 problems, there are still areas
of mathematical recreations that have not been discussed. In deciding
which topics to include, we were guided partly by our desire to
discuss certain mathematical concepts and applications, partly by which
kinds of problems are classic or standard, and partly by our own
preferences. Many other types of problems may be found in the
literature.
For additional problems, we recommend the books appearing in the
bibliography. These and other books of recreational problems may
be found in libraries. (Check the card catalog under Mathematical
Recreations.) To keep in contact with recent developments, we suggest
Martin Gardner's "Mathematical Games" column in current issues of
Scientific American magazine. We also recommend the Journal of
Recreational Mathematics, although some of the articles require more
mathematical background than does this book.

Appendix A
Some Basic Algebraic
Techniques
The ability to solve algebraic equations is one of the most important
of mathematical skills. In this appendix, we review some of the
standard basic techniques:
A. How to solve a linear equation in one unknown.
B. How to solve a linear inequality in one unknown.
C. How to use the quadratic formula to solve a quadratic equation
in one unknown.
D. How to use factoring to solve a quadratic equation in one unknown.
E. How to use the substitution method to solve a simultaneous system
of two linear equations in two unknowns.
F. How to use the method of elimination to solve a simultaneous
system of two linear equations in two unknowns.
G. How to solve a simultaneous system of three linear equations in
three unknowns.
H. How to solve a system of two equations in two unknowns, where
one equation is linear and the other is quadratic.
In our discussion of each of these methods, we give the general
explanation on the left and an example on the right.
321

322
APPENDIX A
A. How to solve a linear equation in one unknown.
Method
The object is to simplify the
equation to the form x = a.
1. Multiply out to remove
parentheses (if there are any).
2. If there are fractions, multiply
both sides of the equation by the
least common denominator. (If
you can't find the least common
denominator, you may multiply
by the product of all the
denominators.)
3. Transpose all x terms to one
side of the equation, and all
constant terms to the other side.
4. Combine the x terms, and
combine the constants.
5. Divide both sides by the
coefficient of x*
6. Check your answer in the
original equation.
Example
3(x ■^2)-^lx = 7(2 - 2x)
3x -f 6 -f fx =
\4x
20(3x -f 6 -f |x = V - 14x)
60x -f 120 + 25x = 56 - 280x
Add 280x to both sides and
subtract 120 from both sides:
60x -f 120 -f 25x + 280x - 120
= 56 - 280x -f 280x - 120
365x = - 64
(Note that there is exactly one
solution.)
Check:
3(i^ + 2) + Ki^)^VB-2(^)]
365
yes — 365 •
* If the coefficient of x should be zero, then either there are no solutions (if the constant
term is not zero) or else every value of x gives a solution (if the equation becomes
Ox = 0). Otherwise, the equation has exaaly one solution.

SOME BASIC ALGEBRAIC TECHNIQUES
323
B. How to solve a linear inequality in one unknown.
Example
3x - ^ < - 4(7 -
Method
The object is to transform the
inequality into the form
X > a or X < a.
1. Solve as if it were an
equation. That is, repeat steps 1
to 5 above. The only difference
is that if you multiply or divide
both sides of an inequality by a
negative number, then the sense
of the inequality is reversed.*
2x)
?>x -\< -28 -f 8x
5(3x - i < -28 -f 8x)
15x - 1 < -140 -f 40x
15x - 1 - 40x -f 1 <
- 140 -f 40x - 40x -f 1*
-25x < -139
(since — 25 is negative)
The answer is all values of x
that are greater than ^.
Note that, in general, there are an infinite number of solutions,
although, as in solving a linear equation—method A above—there
will be no solutions if the inequality reduces to Ox > a where a is
positive or to Ox < a where a is negative.
6. As a check, you can try some
values in the indicated range to
see that they do work, and try
some values not in the indicated
range to see that they don't
work.
6 > ^, therefore 6 should work:
3(6) -l=lli
-4[7 - 2(6)] = 20
17^ < 20.
On the other hand, 5 < ^, so 5
should not work:
3(5) - ^ = 14^
-4[7 - 2(5)] = 12
14^ > 12.
* By properly chcx)sing the side of the
inequality to which x is transposed, we can
avoid having to multiply or divide by a
negative number.
* 15x - 1 < - 140 + 40JC
15jc - 1 + 140 - \5x
< - 140 + 40jc + 140 - \5x
139 <25jc
W < X.

324
APPENDIX A
C. How to use the quadratic
equation in one unknown.
Method
The object is to first transform
the equation to the form
ax^ -i- bx -^ c = 0, and then to
apply the quadratic formula,
which is given below.
1. Multiply out to remove
parentheses (if there are any).
2. Transpose all terms to one
side of the equation.
3. Collect terms and write in
order of descending powers of x.
4. Use the quadratic formula to
solve for x. The formula says
that the solutions of
ax^ -\- bx -\- c = 0 are
and
2a
(If b^ - 4ac is negative, then
there are no solutions. If
b^ - 4ac = 0, then there is one
-b
solution: x = . Otherwise,
2a
there are exactly two solutions.)
-b^^-
- 4ac
2a
-b-^b^-
- 4ac
5. Check your answers in the
original equations.
formula to solve a quadratic
Example
- 2x(x - 4) = 2x' -f 3
- 2x' -f 8x = 2x^ -f 3
0 = 2x^ -f 3 -f 2x2 - 8;c
0 = 4x2 - 8x + 3
a = 4, the coefficient of x^
b =^ —8, the coefficient of x
c = 3, the constant term.
-(-8) + V(-8)2-4(4)(3)
2(4)
± ^64 - 48
±x/l6
±4
There
and
are
X =
X =
exactly
8 + 4
8
8-4
8
two
12
~ ~8
4
~ 8 "
solutions
3
~ 2
1
2'
Check:
-2(1X1-4)^2(1)^+3
-3(^)^2(1)+ 3
¥^! + 3
¥ = ¥•

SOME BASIC ALGEBRAIC TECHNIQUES
325
Also,
-2G)Q-4)^2G)^+3
(-1)(-J)2 20 + 3
^i + 3
^ 7
D. How to use factoring to solve a quadratic equation in one
unknown.
Method
The object is to first transform
the equation to the form
ax^ ■¥ bx ■¥ c = 0^ and then to
factor into a product of two
Hnear factors, and then to set
each factor equal to zero.
Steps 1 through 3 are the same
as in the method above.
4. Factor the nonzero side into a
product of linear factors.
5. Set each of the factors equal
to zero. (If the product of two
numbers is zero, then at least
one of the numbers must be
zero.)
6. Solve each of the linear
equations (as in method A above)
to obtain two solutions.
7. Check, as in method C above.
Example
- 2x{x - 4) = 2jc2 + 3
- 2x^ -f 8x = 2x2 -f 3
0 = 2x2 -f 3 -f 2x2 - g^
0 = 4x2 - 8x + 3
0 = (2x - 3)(2x - 1)
2x
0 or 2x - 1 = 0
2x = 3 or 2x =
X = I or X =
See C for the check.
Note that the method of factoring is usually the simpler way to
solve a quadratic equation, provided that you are able to find the
factors of ax^ ■¥ bx -^ c. If you are not able to do so, then you can
always rely on the quadratic formula.
E. How to use the substitution method to solve a simultaneous
system of two linear equations in two unknowns.
Method
The object is to reduce the
solution of the system to the
solution of one equation in one
Example
(i) X - 143; = -31
(ii) 4x -f 33; = - 6

326
APPENDIX A
unknown by substituting for one
of the unknowns an expression
in terms of the other.
1. Use either of the equations to
express one of the unknowns in
terms of the other.
2. Substitute the expression thus
obtained into the other equation.
3. Solve the resulting equation
as in method A above.*
4. Substitute the value obtained
into the expression found in
step 1. This gives the solution.
5. Check your answer in the
original equations.
Adding I4y to both sides of
equation (i), we obtain
(iii) X = -31 -f 143;.
Substituting —31 -I- 14jv for x in
equation (ii), we obtain
4(-31 -f 143;) -f 33; = -6
-124 -f 56y -f 33; = -6
563; -f 33; = -6 -f 124
593; = 118
y = 2.
Substituting 3; = 2 in equation
(iii)
x= -31 -f 14(2)
= -31+28
= -3.
The solution is x = - 3, 3; = 2.
Check:
(i) -3 - 14(2) = -3-28
i-3l
(ii)4(-3) + 3(2)
12 + 6
^-6.
F. How to use the method of elimination to solve a simultaneous
system of two linear equations in two unknowns.
Method
The object is to reduce the
solution of the system to the
solution of one equation in one
Example
(i) 3x + 43; = 18
(ii) 2x - 53; = - 11
* It is possible that in solving the equation in step 2 that we obtain an equation of the
form Oy = k, where /^ is a nonzero constant. In this case, we say that the system is
inconsistent, and there are no solutions.
It is also possible that we obtain an equation of the form O^; = 0. In this case,
any value of y can be used to obtain a value of x. The equations are said to be
dependent and there are infinitely many solutions.
In all other cases the solution is unique.

SOME BASIC ALGEBRAIC TECHNIQUES
327
unknown by eliminating one of
the variables.
1. Multiply both equations by
the proper numbers so as to
obtain two new equations in
which the coefficients of x (or
of 3^) are the negatives of each
other.
2. Add the resulting equations to
eliminate x (or y) and obtain an
equation in one unknown.*
3. Solve as in method A above.
4. Substitute into either original
equation to solve for the other
unknown.
5. Check your answer in the
original equations.
Multiply (i) by 2 and (ii) by - 3
2(3x -f 43; = 18)
-3(2x - 5y = -11)
to obtain
(iii) 6x -\- Sy = 36
(iv) -6x+ 153; = 33.
Adding,
233; = 69
3^ = 3.
Substitute in equation (i),
3x -f 4(3) = 18
3x -f 12 = 18
3x = 6
x = 2.
The solution is x = 2, 3; = 3.
Check:
(i) 3(2) + 4(3) = 6 -f 12 i 18
(ii) 2(2) - 5(3) = 4 - 15 i -11.
G. How to solve a simultaneous system of three linear equations
in three unknowns.
Method
The object is to transform the
system of equations to the form
ax + by -\- cz = d
ey -^ fz = g
kz = m
and then to solve from the
bottom up.
1. Rearrange the equations, if
Example
(i) 3x + 7y -4z = -W
(ii) 2x - 43; -f z = -5
(iii) -5x -\- 2y - Sz = - 12
* If, when we add, the equation becomes Ox + Oy = a, where a # 0, then the system
is inconsistent and there are no solutions. If we obtain Ox -\- Oy = 0, then the equations
are dependent and there are infinitely many solutions. In all other cases, there is a unique
solution.

328
APPENDIX A
necessary, so that x has a
nonzero coefficient in the first
equation.
2. If X has a nonzero coefficient
in any of the other equations,
use method F above to eliminate
X from these equations.
You obtain a new system of the
form
ax -\- by -\- cz = d
ey -^ fz = g
hy -f iz = j.
3. Use the last two equations to
eliminate y from the last
equation.
The system has now been
transformed into the desired
form.
4. Solve the last equation for the
third unknown, as in method A
above.
Use (i) and (ii) to eliminate x
from (ii)
2(3x -f 73; -4^ = -11)
- 3(2x - 43; -f z= -5)
6x -f 143; - 82: = -22
-6x -f 123; - 3^ = 15
(iv)
263; - Uz = -7.
Use (i) and (iii) to eliminate x
from (iii)
5(3x + ly -^z= -11)
3(-5x-f 23; -82 = -12)
\5x -f 353; - 20z = -55
-\5x -f 63; - 242 = -36
(V)
4I3; - 442: = -91.
(i) 3x + 73; - 42 = - 11
(iv) 263; - II2 = -7
(v) 4I3; - 442: = -91
Use (iv) and (v) to eliminate y
from (v)
41(263; - \\z= -7)
-26(413; -44^ = -91)
41(26)3; - 45I2 = -287
-41(26)3; + 11442: = 2366
(vi) 6932 = 2079.
The system becomes
(i) 3x + 73; - Az= -n
(iv) 26y - \\z = -1
(vi) 6932 = 2079.
6932 = 2079
jr — 2Q79 — ^
^ ~ 693 ~ ^

SOME BASIC ALGEBRAIC TECHNIQUES
329
5. Substitute this value into the
middle equation and solve for
the second unknown.
6. Use the values found for both
of the unknowns to find the first
unknown by substituting into
the first equation.
7. Check your answer in the
original equations.
263; - 11(3) = -1
263; - 33 = - 7
263; = - 7 -f 33 = 26
y= 1
3x -f 7(1) - 4(3) = -11
3x -f 7 - 12 = -11
3x = -11 - 7 -f 12 = -6
x= -2
The solution is
X = -2, y = \i z = 3.
Check:
(i)3(-2) + 7(l)-4(3)
= -6 -f 7 - 12 = -11
(ii)2(-2)-4(l) + (3)
= -4 -4 -f 3 = -5
(iii) -5(-2) + 2(l)-8(3)
= 10 -f 2 - 24 = - 12.
Note that, again, it is possible for the equations to be inconsistent
[if we obtain an equation 0 = a (^ 0)] or dependent (if an equation
becomes 0 = 0), in which cases there will be no solutions or infinitely
many solutions respectively. In all other cases, there will be a unique
solution.
An alternative approach to solving a system of three linear equations
in three unknowns is to use the method of substitution (method E
above) to reduce the system to two equations in two unknowns, and
then proceed as in method E or F.
H. How to solve a system of two equations in two unknowns,
where one equation is linear and the other is quadratic.
Method
The object is to reduce the
system to one quadratic equation
in one unknown and then to
solve as in method C or D above.
1. Use the linear equation to
solve for one of the unknowns
in terms of the other.
Example
(i) x^ -f 2x - y = - 1
(ii) 2x -y =1
Use (ii) to solve for y in terms
of X

330
APPENDIX A
2. Substitute into the quadratic
equation.
3. Simplify and solve as in
method C or D.
Note that there will be zero,
one, or two solutions.
4. Use the equation in step 1 to
find the other unknown.
5. Check your answer in the
original equations.
2x -y = l
2x = 1 -¥ y
y = 2x-l.
Substitute into (i)
x" + 2x-{2x- ly = -1.
x^ + 2x - (4x2 - 28x -f 49) = - 1
x^ -f 2x - 4x' + 28x - 49 = - 1
- 3x^ -f 30x - 49 = - 1
0 = 3x2 - 30x -f 48
0 = 3(x2 - lOx + 16)
0 = x^ - lOx + 16
0 = (x - 2)(x - 8)
X = 2 or X = 8
3; = 2x - 7
y = 2(2) -1 = -3
or y = 2(8) -7 = 9
There are two solutions:
X = 2 and jy = — 3
or X = 8 and y = 9.
Check:
(i) 22+ 2(2)-(-3)2
=4+4-9
1
(ii) 2(2) - (- 3) = 4 H- 3 = 7.
(i) 82 -f 2(8) - 92
= 64 + 16 - 81 = -1
(ii) 2(8) - 9 = 16 - 9 :^ 7.
PRACTICE
PROBLEMS A
A.l. [a] Solve for the unknowns:
(a) x(x - 1) = 6
(c) x^ + 3x - 7 = x(x - 5)
(e) 3x - 4 < 7x + 1
(b) 4(x - i) + 3x = 2x - ^
(d) 2x^ - 9x + 1 = 0
(0 2x2 - 4x + 3 = 0

SOME BASIC ALGEBRAIC TECHNIQUES 331
(g) 3x -^4y = 12 (h) x^ -f y = 25
2^ + 33; = 7 X - y = -\
(i) 3x - 4y -\- z = 20
X + ly -?>z = - 35
33; - 22 = - 19
A.2. Solve for the unknowns:
(a) lx-Ay= -9 (b) x^ - 8x + 16 = 0
2x + 6y = 1
(c) x(3x - 1) = 2x^ - 7x + 1 (d) X + y + z = ^
2x + y -j- z = 6
3x + 2y - z = I
(e) 7x + 8 > i - 2x (0 3x - 7(1 - x) = \0x + 2
(g)x^-y = 9
X + 2j; = 13

Appendix B
HAT Does mTV
LIKE ?^>/
M
Mathematical
Induction
Often in life, when we observe that some relationship has held true
up to now, we conclude that it will continue to hold in the future.
For example, even people who know nothing of the physical laws
governing the motion of heavenly bodies "know" that the sun will rise
tomorrow morning, because it has risen every morning in the past.
Such reasoning is called inductive reasoning.
Although inductive reasoning plays a very important role in the
physical and natural sciences, it can lead to mistakes. Consider the
expression
x' - lOx^ -f 35x' - 50x + 24.
Observe that if we substitute the number 1, 2, 3, or 4 for x, the value
of the expression becomes zero. We might conclude, by inductive
reasoning, that the expression above will become zero no matter what
integral value we substitute for x. But when we substitute 5, the value
of the expression becomes 24, and so our conclusion is false.
You might argue that our induction is based on only four
observations, hardly enough to draw any conclusions.
But, what is enough? Twenty? Forty?
Consider the expression
jc^ -f X + 41.
When we substitute any integer from 1 to 20 for x, the value of this
expression is a prime number. (See Chapter 4 for the definition of
prime numbers.) In fact, the value of the expression is a prime when-
332

MATHEMATICAL INDUCTION
333
ever any integer from 1 to 40 is substituted for x. Can we conclude
that the value of the expression will be prime no matter what integer
is substituted for x ? No. As a matter of fact, when x = 41, the expression
takes the value 1763, which is not prime.
Does this mean that, no matter how many observations we make,
we still cannot conclude that a result is true in general? The answer
to this question is yes—observation alone, no matter how extensive,
does not prove anything. If we observe that
1 -f 2 + 22 -f 2^ +
-f 2":
1
for each value of w, 1 <n< 100 or 1 < w < 1000, it is still conceivably
possible that the result will fail to hold for some larger values of n.
Since we cannot make infinitely many observations, how can we
hope to prove that the result above holds for all values of w ? The answer
lies in the method of mathematical induction.
The principle of mathematical induction is one of the defining
properties of the positive integers. Its statement is as follows:
Let S„ be a statement about a general positive integer n. If
(i) Sj is true (that is, the statement is true when n = \)
and
(ii) whenever Sk is true, then Sk+ \ is also true (that is, for each
positive integer k^ the truth of Sk implies the truth of Sk+ i),
then S„ is true for all positive integers n.
That is, to prove that a statement holds for all positive integers, it
is enough to show that it holds for 1 and that whenever it holds for a
positive integer k^ it also holds iot k + 1.
A heuristic argument justifying the principle of mathematical
induction is as follows: By (i), Sj is true. Therefore, by (ii), S2 must
also be true (since Si is). But then, again by (ii), S3 must be true
(since S2 is). Etc. Hence, S„ must be true for all positive integers.
Using a similar argument, we see that it is not always necessary to
start our induction at the number 1. If we wish to prove that some
statement S„ holds true for all integers n greater than or equal to five,
we first show that S5 is true, and then we show that Sk implies Sk + 1
for all positive integers k>5.
Let us now illustrate the use of the principle of mathematical
induction. We use it to prove that, for every positive integer w,
1 + 2 + 2^ + • • • + 2" = 2" + • - 1. We present the proof on the right
below, and explain the method on the left.
Given S„,
we wish to show that S„ is true
for all positive integers.
We must first show that Si is
true.
S„: 1 -f 2 + ••• + 2" = 2"-^' - 1
We wish to prove that S„ is true
for all positive integers.
(i) S,: 1 -f 2 = 2^ - 1
This is certainly true (3 = 3).

334
APPENDIX B
We next assume that Sk is
true for some general positive
integer k and, using this fact,
we try to show somehow (using
any tools of algebra, logic, etc.)
that Sk + 1 must also be true.
(Note that S;ij + i is obtained by
replacing n with k + 1 in S„.)
If we succeed, then we can
conclude that S„ is true for all
positive integers n.
(ii) Assume Sk is true for some
general positive integer k:
1 +2 + ••• +2* = 2*+i - 1 (1)
We try to show that Sk + i must
then also be true. (Note,
Sa + 1: 1 + 2 + • • • + 2*-f 2* + 1
= 2* + 2 _ 1.)
Add 2* + 1 to both sides of
equation (1).
1 + 2 + • • • + 2* + 2* + 1
= 2*+i - 1 + 2*+i
= 2 • 2* + 1 - 1
= 2* + 2 _ 1
But this is just S;ij + i. Hence, by
the principle of mathematical
induction, S„ is true for all
positive integers n.
We conclude this appendix with one more example of an inductive
proof.
To present the result that we are about to prove, we introduce the
symbol " !" (read "factorial")-
By w!, we mean the product of all the integers from 1 to n. That is,
n\ = n(n- 1)(« - 2) • ... • 3 • 2 • 1.
For example,
4! = 4 • 3 • 2
Note that
(n + 1)! = (w+ \)n(n - 1) • ..,
We are now ready to state the theorem we wish to prove.
Theorem For every integer w > 4, 2" < w!. That is,
S„:2''<w!, for w > 4.
Since 4 is the smallest value for which S„ is claimed to be true,
we start the induction at w = 4.
S4: 2^ < 4! is true since 16 < 24.
Assume Sk is true. That is, 2* < k\.
(We must show that 8,^ + 1 is true. That is, 2* + 1 < (/j + 1)!.)
1 = 24.
3 • 2 • 1 = (« + 1) • «!.

MATHEMATICAL INDUCTION
335
Note that 2 < /j + 1, since k>4.
Hence,
2 - 2f^ <(k+ \) - 2^^ <(k-^ \) ■ kl (since Sk is true).
Thus,
2k+\ <^k+ 1)!
and so S/t + 1 is true. The proof is now complete.
B.l. [s] Prove by mathematical induction:
n(n + 1)
1 + 2 + 3 +
-\- n =
B.2. Prove by mathematical induction:
F + 2^ + 3^ + • • • + w^ = ^
6
B.3. [s] Consider a rectangle and n straight line segments, each of
which begins on one edge of the rectangle and ends on another
edge. Prove, by mathematical induction, that the resulting map
can be colored using exactly two colors, so that bordering regions
are of different colors. (See Chapter 6, page 198, for a discussion
of map coloring.)
B.4. Use mathematical induction to prove that at least n + 1 integral
weights are needed to be able to balance any integral weight not
exceeding 2" grams. (See Chapter 5, page 152.)
B.5. Prove by mathematical induction that there are 2" cases in the
truth table for a statement involving n distinct variables:
Pi > P2 > . . . 5 P«.
B.6. Prove by mathematical induction that n guesses will suffice to
determine a previously chosen positive integer that is between
1 and 2" - 1, provided that, with each incorrect guess, the guesser
is told whether the chosen number is larger or smaller than the
guess. (See Exercise 5.16.)
PRACTICE
PROBLEMS B

Appendix C
Probability
YOV/R CHANCES OP
KOLLl/V/O At)0VA8L5
SIX /)RE5u;rf smU-.
I'LL VOi^dl-E.

PROBABILITY 337
Although there are many games whose outcomes are completely determined by the skill of the
players (see Chapter Seven), there are many other games in which chance is also an important factor. It
was in considering certain gambling aspects of such games that Pascal and Fermat (in the seventeenth
century) laid the foundation for the formal study of probability. (Cardano had done some work in the field
earlier, but it is really Pascal and Fermat who developed the subject.)
According to Eves/'^ Chevalier de Mere, a gambler of the time, proposed the following problem to
Pascal in 1654. (It should be noted that this problem had been posed in the literature as early as 1494 in
Pacioli's Suma.)
Two equally skilled players are playing a game of chance where, at each play, one of the players
wins a point. The game is to end when one player attains n points and, as a resuh, wins the total "pot." If
the players must stop playing before either wins (e.g., when A has a points and B has b points), how should
the money be distributed between the two players?
The general study of probability goes far beyond the scope of this book. However, we present a
limited view of the subject which suffices to study most games of chance as well as certain other
recreational problems. To introduce this view, we consider the following sample problems. Note that
Problem C.6b is really a special case of the problem of Chevalier de Mere from which probability theory
stems.
SAMPLE PROBLEMS
Problem C.l
Fenwick Featherhead has trouble making decisions, so he always carries around a pair of dice (one red and
one white) to help him out. Today, at lunch, Fenwick had difficulty deciding whether to have a chocolate
malted or a banana split for dessert. Taking out his trusty dice, he said, "I'll roll both dice. If either shows
a six, I'll buy a malted; if the sum of the dice is either seven or eleven, I'll order the banana split; if both
occur, I'll have two desserts; otherwise, I'll have no dessert at all." Assuming that the dice are honest, what
is the probability that
a) Fenwick has a banana split?
b) Fenwick has a malted?
c) Fenwick has both a malted and a banana split?
d) Fenwick has no dessert at all?
Problem C.2
Before the era of the internet, fax machines, etc., all of the three-digit area codes used by the telephone
companies in the United States satisfied the following rules:
a) The first digit could not be 0 or 1.
b) The second digit had to be 0 or 1.
c) The third digit could not be 0.
d) The third digit could be 1 only if the second digit is 0. How many possible area codes were
there?
Problem C.3
At a track meet in which six teams participated, each team entered three runners in the one-mile run.
Medals were awarded to the first three finishers: Gold for first, silver for second, and bronze for third.
a) Assuming that there were no ties, in how many different ways could the medals have been
distributed among the runners?
b) If all eighteen runners are evenly matched, what is the probability that the three medal winners
all belong to the same team?
(^^: H. Eves, An Introduction to the History of Mathematics. Rinehart, New York, 1960.

338 APPENDIX C
Problem C.4
You are "the shooter" in a game of Craps, which is being played according to the following rules: You roll
two dice. If the total is seven or eleven you win immediately; if the total is two, three, or twelve, you lose
immediately; if the total is any other number (referred to as "your point"), you continue rolling the dice
until you obtain either a seven, in which case you lose, or your point, in which case you win.
a) What is the probability that you win?
b) Before rolling the dice, you place a bet. If you lose the game, then you lose the amount of the
bet; if you win the game, then the amount of your winnings is determined as follows:
If you win by rolling a seven or eleven, you win the amount of your bet (even odds);
If your point is six or eight and you win, then again you win the amount of your bet (even odds);
But, if your point is four or ten and you win, you win twice the amount of your bet (odds of two to
one); And, if your point is five or nine and you win, you win three halves times your bet (odds of
three to two).
What is the expected value of this game for you, if you make a bet of $1 each time you play? (I.e.,
on the average, how much will you win or lose per game if you play the game many times?)
Problem C.5 [based on 9, Chapter 3, Problem 8]
Captain Damon "dead-eye" Dimwitty, the famed destroyer commander of World War II, was
noted for his uncanny ability to sink enemy submarines. Based on a survey of his battle experiences, it was
computed that each depth charge he fired had probability 1/2 of scoring a direct hit and thus sinking its
target. It was also found that when a depth charge did not result in a sinking, it nevertheless caused less
severe damage to an enemy vessel with probability 1/4 and that two damaging explosions are sufficient to
sink a submarine. Finally, and this is the remarkable part, Captain Dimwitty only missed his adversary 1/4
of the time.
With these facts in mind, what is the probability that Captain Dimwitty would be able to sink an
enemy sub by using no more than four depth charges?
Problem C.6
Two firemen, A and B, are playing the following game. Each antes up a certain amount of money
to form a pot. A die is rolled. If the result is a 1 or a 6, A gets one point; otherwise B gets one point. The
first player to attain four points wins the game and takes the money that is in the pot.
a) If they wish there to be $100 in the pot, what is the amount to the nearest penny that each
should put in at the beginning, so that the game is a fair game (i.e., expectation = 0)?
b) After they have been playing for a while, the fire bell rings, and they must interrupt the game.
If at this point, A is leading two points to one, how should the $100 be divided equitably between
them?
> WHAT IS PROBABILITY?
Suppose we are "conducting an experiment" (such as playing a game of chance) in which a
number of different outcomes are possible. If some of these outcomes are viewed as favorable and others
are viewed as unfavorable, we may speak of the probability, or likelihood, that a favorable outcome will
occur. For example, if we are about to roll two dice and wonder about the likelihood that at least one of the
dice will show a six^^\ then any outcome in which at least one die shows a six will be counted as favorable,
and all the other outcomes will be considered as unfavorable. Alternatively, if we are interested in the
probability that the sum of the numbers showing on the two dice will be seven, then any outcome for which
(2): When we speak of the number showing on a die, we mean on the top face of a die (unless otherwise
indicated); similarly, when we speak of the sum of two or more dice, we refer to the sum of the numbers on
the top faces of the dice.

PROBABILITY 339
the sum of the two dice is seven will be viewed as favorable, and any other outcome will be regarded as
unfavorable.
We sometimes refer to any particular subset of the set, S, of possible outcomes of an experiment
as an event. Thus, in the example above, we may speak of the event that at least one die shows a six, or of
the event that the sum of the dice is seven, and so on.
PRACTICE PROBLEM SET C.A
1. 0 a)If we toss a penny, a nickel, and a dime, what are the eight possible outcomes (considering heads
and tails assuming a coin does not land on its edge)?
b) List the outcomes in each of the following events:
i) The penny comes up heads.
ii) The nickel and the dime come up the same.
iii) There are more heads than tails.
2. a) If we toss a penny, a nickel, a dime and a quarter, what are the sixteen possible outcomes?
b) List the outcomes in each of the following events:
i) The dime comes up tails,
ii) Exactly 15 cents comes up heads,
iii) At least 15 cents comes up heads,
iv) There are more heads than tails.
3. ^ a) As an experiment, you administer a four question true-false examination to a subject. If you
consider each answer to be either right or wrong, find two different outcome sets which could be
used to represent the possible outcomes of the experiment.
b) For each of these outcome sets, list the outcomes in each of the following events:
i) The subject gets 100% correct,
ii) The subject gets 75% correct,
iii) The subject gets 50% correct.
In this chapter, we will be concerned mainly with experiments for which there are only a finite
number, «, of possible outcomes, all of which are equally likely.^^^
However, let us first digress to point out that this is not the most general situation. If three horses
are entered in a race, there is no reason to assume that each of the horses is equally likely to win. One
horse is usually the favorite, meaning that most people estimate its likelihood of winning to be greater than
that of any other horse.
In general, given the finite outcome set S = {5;,5j,...,5„}, each outcome, 5,, may be assigned a
nonnegative weight w, (i.e., w, > 0 for all /), called the probability of 5„ such that the sum of these assigned
n
weights is 1. (I.e., Xw^^l) The probability of an event E is then the sum of the probabilities of the
/■=!
outcomes in the set E.
What weights should be assigned to each of the outcomes is not always an easy question. If Fleet
of Foot, Runs Like the Devil, and Ton of Lead are racing each other, Benny the Bookie might decide to
assign them weights of 3/4, 1/4, and 0 respectively. That is, he may feel that the probability that Fleet of
Foot may win is 3/4, that the probability of Runs Like the Devil winning is 1/4, and Ton of Lead has no
chance at all. Hardluck Harry may disagree, and he may assign the weights 5/12, 5/12, and 1/6
respectively.
In some situations, theoretical considerations and/or empirical evidence dictate what weights we
should assign. Given an ordinary coin, there is no reason to suspect that, if it is flipped, a head is any more
likely to appear than is a tail; we should think that the two outcomes are equally likely. This can be borne
out by experimental observation. Hence, we usually assign the two outcomes equal weights. In general, if
an experiment has n possible outcomes, all of which we consider, for one reason or another, to be equally
likely, then all n outcomes should be assigned the same weight. Since the sum of the weights of all
(3): Roughly speaking, this means that, if the experiment were repeated a very large number of times, all
possible outcomes would occur with approximately the same relative frequency.

340 APPENDIX C
outcomes must be 1, we must assign the weight l/n to each outcome. Thus, for example, if we roll a die, we
usually assign a probability of 1/6 to each of the six possible outcomes. We indicate this by saying that we
roll an "honest" die. Similarly, we speak of a "fair" coin, if the probability of a head and the probability of
a tail are both 1/2.
Since the probability of an event should be the sum of the probabilities of the outcomes in that
event, we make the following definition:
Definition C.l: If/of the possible outcomes of an experiment are favorable for a particular event E and if
the experiment has n possible outcomes which are equally likely, then the probability that the event E will
occur (referred to as the probability of £) is given by
^ _ / the number of outcomes favorablefor E , _
Pr(£) =^ = (C.l)
n the total number of possibleoutcomes
Note that this is consistent with our assigning each element of the outcome set a weight of l/«.
For example, when we roll a die, the set of possible outcomes is {1,2,3,4,5,6}. Therefore,
i) the probability of rolling a number greater than four is 2/6 = 1/3 (two outcomes, 5 and 6, are
favorable for this event);
ii) the probability of rolling a number less than seven is 6/6 = 1 (all outcomes are favorable);
iii) the probability of rolling a number larger than six is 0/6 = 0 (no outcomes are favorable).
Note that, in general, 0 </< «, so that
0< —= pr(£)<l
n
The only way that pr(£) could be 0 is if/= 0; i.e., no outcomes are favorable for E, and so E is
impossible^'*^ Similarly, if pr(£) = 1, then/= «, and so all outcomes are favorable for E. Thus, E is certain.
In any case, pr(£) cannot be less than zero (i.e., negative) or greater than 1.
Note also, that if E is any event for which f outcomes are favorable, and if E' is the complement of £" (i.e.,
the event that E does not occur), then ^-/outcomes are favorable for E'. Hence,
PT{E) = ^^^=--1= 1 -pr(£) (C.2)
n n n
Therefore, to compute the probability of an event, it is possible (and sometimes easier) to compute
the probability of its complement and then to use formula (C.2).
Referring to the example of rolling one die, the probability of obtaining a number that does not
exceed four is equal to 1 minus the probability of obtaining a number greater than four; i.e., it is 1 - (1/3) =
2/3. (Note that four of the six possible outcomes are favorable.)
PRACTICE PROBLEM SET C.B
1. 0 If a fair penny, nickel, and dime are tossed, what is the probability of each of the following events?
a) The penny comes up heads.
b) The nickel and the dime come up the same.
c) There are more heads than tails.
^"^^: Sometimes the probability of an event is so small as to make it effectively zero for most practical
purposes. Nevertheless, a favorable outcome is possible, although extremely unlikely. We often say that
the event in question is "virtually impossible." A real life example of this occurs when the weather bureau
says that the probability of rain is zero. As we have learned from experience, this does not guarantee that it
will not rain. A special case of "effectively zero" occurs when there is an infinite number of equally likely
possible outcomes and only one or two (or any finite number) are favorable for a particular event E. Then
the probability of that event is less than any positive number 8 , and hence may be said to be zero. In such
a case, E is far less likely than the weather bureaus "no chance of rain"; but, even so, as a matter of
principle, E is not absolutely impossible. We may be justified in betting heavily against such an event, but,
practically speaking, the stake can be lost.

PROBABILITY 341
d) All three coins come up heads.
2. If you roll an honest die, what is the probability of each of the following events?
a) An odd number is obtained.
b) An odd number greater than three is obtained.
c) The number obtained is less than two.
3. 0 If £■ is an event and pr(£) = 1/8, find pr(£").
> ODDS
Another way of expressing a statement about the probability of an event occurring is in terms of
odds. Given an equally likely outcome set, if twice as many outcomes are favorable for an event to occur as
are unfavorable, we frequently say that the odds are 2 to 1 in favor of the event. In general, given an equally
likely outcome set, if f outcomes are favorable and «-/outcomes are unfavorable, we say the odds are/to
n-frn favor of the event occurring, or that they are n-f to/against the event occurring.
We frequently express this fact by writing the odds in favor of E as the ratio f: n - f, or as the
fraction . Thus, for example, odds of 4 : 2 (4/2) are the same as odds of 2 : 1 (2/1).
n-f
In terms of probability, pr(£) = — and pr (^^ = , so the odds in favor of £ are
n n
f /„ pr(£) pr(£)
n-f ("-/)/ pr(F) l-pr(£)
/ n
This suggests the following definition:
Definition C.2: Up represents the probability that an event E occurs, then the odds in favor ofE are/? to
1 - p and the odds against E arel-p to p.
If we roll an honest die, the odds against getting a number greater than four are 2 : 1, since four
outcomes are unfavorable and only two are favorable. Note that the probability of getting a number greater
,Jl/^ 2/
than four is 1/3 and —.— = —7 = —
1/ 1/ 1
/3 /3
PRACTICE PROBLEM SET C. C
1. 0 If we roll an honest die,
a) What are the odds in favor of getting a three?
b) What are the odds against getting a three?
2. On an honest roulette wheel, there are thirty-eight equally likely possible outcomes: 00, 0,1, 2, 3, 4,...,
32, 33, 34, 35, 36.
a) What are the odds against getting an odd outcome?
b) What are the odds against getting an outcome between 13 and 24 inclusive?
c) What are the odds in favor of getting a particular outcome, say 2?
Usually, mention of "odds" arises in connection with betting. If we are offered odds of r to s on a
bet, this means that we will have to pay $s if we lose the bet, but we will receive $r if we win. This may not
reflect the "true odds" of the bet as defined above. We will return in the section on expectation to discuss
how the two uses of the word "odds" are related.

342 APPENDIX C
> EQUALLY LIKELY OUTCOMES
We emphasize that formula (C.l) for calculating probability is applicable only if all possible
outcomes of the experiment are equally likely/^^
Determining a set of equally likely outcomes may sometimes require careful thought.
For example, suppose we throw two honest dice and ask, "what is the probability of getting a seven?" We
might reason as follows: There are eleven possible outcomes (2,3,4,5,6,7,8,9,10,11, and 12), one of which,
7, is favorable; therefore, the probability of getting a seven is 1/11. Wrong! The reasoning is faulty. The
eleven possible outcomes are not equally likely; in the long run, for example, 7 will appear more often than
2 (as we will see below).
The following attempt is better, but is still faulty: There are twenty-one possible outcomes ( 1 and
1, 1 and 2, 1 and 3, 1 and 4, 1 and 5, 1 and 6, 2 and 2, 2 and 3, 2 and 4, 2 and 5, 2 and 6, 3 and 3, 3 and 4, 3
and 5, 3 and 6, 4 and 4, 4 and 5, 4 and 6, 5 and 5, 5 and 6, and 6 and 6), three of which are favorable ( 1 and
6, 2 and 5, and 3 and 4); so the probability of getting a seven is 3/21 = 1/7. Wrong again!
The mistake here is again that the twenty-one possible outcomes are not equally likely. For
example, 1 and 2 will appear twice as often as 1 and 1. To see this, suppose one die is red and the other is
white. Then 1 and 1 will occur only if both dice have 1, whereas 1 and 2 will occur if the red die shows 1
and the white one 2, or if the red shows 2 and the white 1.
This then leads us to the proper solution of the problem. If we consider the dice as being red and
white, then there are actually 36 possible outcomes, which are equally likely. If we let {r,w) represent the
outcome in which the red die shows r and the white die w, then the possible outcomes are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), and (6,6).
Of these, six are favorable ( (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) ). Thus, the probability of
getting a seven is 6/36 = 1/6.
(Note that, now that we have found the equally likely outcome set, we know what weights we
should assign if we want to use the outcome set {2,3,4,..., 12}, where the outcomes represent the sum of the
dice. The outcome 2 should be assigned a weight of 1/36, since one of the thirty-six possible outcomes
results in a sum of 2; the outcome 3 should be assigned the weight 2/36; etc.)
PRACTICE PROBLEM SET CD
1. 0 a) Toss three pennies; each lands heads or tails. What equally likely outcome set should we consider
if we are interested in the probability that exactly two of the coins come up heads? What is the
probability in question?
b) What weights should be assigned if we wish to use the outcome set {0 heads, 1 head, 2 heads, 3
heads}?
^^^: However, probability theory can also handle the case of outcomes that are not equally likely, as in the
example of the horse race. Although our discussion of the formulas and theorems presented in the
remainder of the chapter will primarily deal with the equally likely outcome case, all formulas and
theorems of the chapter hold for a more general discussion of probability (unless otherwise indicated as in
the case of (C.l)). They may be applied in this broader context to some of the problems and exercises of
the chapter, provided that we are told the underlying weights or probabilities to use. (See, for example, the
solution of Problem C.5, later in the chapter.)

PROBABILITY 343
2. Two possible outcome sets for Problem 3 of Practice Problem Set 9.1 are: {(w,w,w,w), (w,w,w,r),...,
(r,r,r,r) }, where w stands for "wrong" and r stands for "right", and {0,1,2,3,4}, where each outcome
represents the number of correct answers.
a) Assuming that the subject guesses randomly at each answer, which of the above sets contains
equally likely outcomes?
b) What weights should be assigned to the outcomes in the other outcome set?
3. 0 Randomly draw a card from an ordinary 52 cards deck of playing cards. What equally likely
outcome set should we use if we are interested in the probability that the chosen card is a picture card?
What is the probability in question?
The example above points out one of the difficulties in computing probabilities of the type we
consider - expressing the results of the experiment in terms of outcomes that are equally likely.
Once this has been done, the next difficulty is counting the number of possible outcomes and the
number of favorable ones.
You now should be able to solve Problem C.l. Try it again if you haven't already solved it.
SOLUTION TO PROBLEM C.I
In an experiment such as the one considered above, we may actually list all possible outcomes and
count how many there are and which are favorable.
For example, in Problem C. 1, if Eg is the event that a seven or eleven occurs, then
£a = {(1,6),(2,5),(3,4),(4,3),(5,2),(5,6),(6,1),(6,5)} , so pr(E,) = 8/36 = 2/9, and Fenwick
has a banana split with probability 2/9. Similarly, if Eb is the event that at least one die shows a six, then
£b={(l,6),(2,6),(3,6),(4,6),(5,6),(6,l),(6,2),(6,3),(6,4),(6,5),(6,6)}.
Thus prC^b) = 11/36, and so Fenwick has a malted with probability 11/36.
The probability that Fenwick has both a malted and a banana split can be computed in a similar
manner:
pr(£e) = pr({(l,6),(5,6),(6,l),(6,5)})= 4/36 = 1/9 .
To compute the probability that Fenwick has no dessert, it is easier to first compute the probability
of the complementary event (that he has at least one dessert) and then to subtract from 1.
£d'={(l,6),(2,5),(2,6),(3,4),(3,6),(4,3),(4,6),(5,2),(5,6),(6,l),(6,2),(6,3),(6,4X
so pr(£d') = 15/36 = 5/12, and pr(£^) = 1 - 5 /12 = 7/12.
> COUNTING: ADDITIVITY
In many experiments, there are just too many possibilities to list, and so we must develop other
techniques for counting. One helpful technique is to partition the outcomes into disjoint cases (i.e., cases
which have nothing in common).
Reconsider Problem C.la. As we saw above, there are 36 possible outcomes when two dice are
thrown. We must count how many of these outcomes result in a seven or eleven. The answer is eight - six
which resuh in a 7 (as we saw above) and two which result in an 11 ((5,6) and (6,5)). In other words, the
event E^, that a seven or an eleven occurs, may be broken up into two disjoint events - Ei, that a seven
occurs, and Ei, that an eleven occurs. The number / of outcomes favorable for E^ is the sum of the
number,/,, of outcomes favorable for E\, and the number,y^, of outcomes favorable for Ej. In terms of
probabilities, pr(£'a) =fln = (fj+fiVn = {fxin) + {filn) = pr(£'i) + pr(£'2). This holds whenever E can be
broken up into disjoint events Ei and Ej.
We state this result as a theorem.

344 APPENDIX C
Theorem C.l: If the event E is made up of two disjoint events E\ and £"2, then
pr(£) = pr(£,) + pr(^2).^'^ (C.3)
Thus, in Problem C.l a,
pr(E,) = 6/36 + 2/36 = 8/36 = 2/9 .
We emphasize the importance in formula (C.3) of the events Ei and E2 being disjoint. In Problem
C.lb, if £"1 is the event that the red die shows six, and E2 is the event that the white die shows six, then we
might be tempted to say that pr(£'b) = prC^*!) + pr(£'2). Wrong!
E, = {(6, 1),(6,2),(6,3),(6,4),(6,5),(6,6)} ,
so
pr(£:,) = 6/36 = 1/6;
and
£2={(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)},
so
pr(£:2) = 6/36 =1/6;
however, as we saw above,
pr(£b) = 11/36.
Obviously, pr(£'b) = 11/36 ^ 1/6 + 1/6 = pr(£'i) + pr(£'2). The reason for the error is that E\ and £"2 are not
disjoint; the outcome (6,6) belongs to both and hence is counted twice in pr(£'i) + pr(£'2).
In the case that E\ and E2 are not disjoint, we can compensate for our error by subtracting off
outcomes that are counted twice. That is, we have the following theorem.
Theorem C,2: UE\ and £"2 are events associated with the same experiment, and if E is the event consisting
of those outcomes which belong to E\ or E2 (or both), then
pr(£) = pr(£,) + pr(E2) - pr(£, ^£2)
where £"1 n E2 (called the intersection of £"1 and E2) consists of those outcomes which are in both Ei and £"2-
In Problem C. lb, pr(£:,) = 1 / 6, pr(£:2) = 1 / 6, and pr(£:,n £2) = 1 / 36, so
pr(£:b) =1/6 + 1/6-1/36 = 11/36 .
If we are careful, we can always choose the manner in which we break up an event so that the
smaller events are disjoint. For example, in Problem C. lb, let E\ be the event that the red die shows six but
the white die does not; let E2 be the event that the white die shows six but the red die does not; and let £3 be
the event that both dice show six. Then £"1, £2, and £3 are pairwise disjoint (i.e., no two have any outcomes
in common) and together form £. There are five outcomes in E\ ((6,1),(6,2),(6,3),(6,4) and (6,5)), five
outcomes in £2 ((1,6),(2,6),(3,6),(4,6) and (5,6)), and one in £3 ((6,6)). Thus
pr(£b) = pr(£,) + pr(£2) + pr(£3) = 5/36 + 5/36 + 1/36 = 11/36.
PRACTICE PROBLEM SET C.E
1.0 If a card is randomly selected from an ordinary fifty-two card playing deck, what is the probability of
each of the following events?
a) The selected card is red.
b) The selected card is a picture card.
c) The selected card is a red picture card.
d) The selected card is red, but not a picture card.
e) The selected card is either red or a picture card.
Obtain the answer to e) in two different ways.
^^^: We remind you once more that, although Theorem C.l and the remaining theorems of this chapter are
only being presented for the equally likely outcome definition of probability, corresponding theorems hold
for the more general definition of probability.

PROBABILITY 345
2. If four fair coins (one penny, two nickels and a dime) are tossed, what is the probabihty that
a) the dime will land on heads?
b) both nickels will land on heads?
c) the value of the coins which land on heads is at least ten cents?
> THE MULTIPLICATION PRINCIPLE
Another helpful technique of counting is based on the Multiplication Principle, which was
discussed in Chapter One. We restate it here to refresh your memory.
Theorem C.3 (The Multiplication Principle): Suppose that a task can be broken up into two steps, the first
of which can be done in m ways and the second of which can be done in n ways (regardless of the result of
the first step). The original task can be done in mn ways.
In the experiment of rolling two dice, we listed all possible outcomes and found that there were 36
in all. An alternate route to this same result would be the following: Suppose that, instead of rolling both
dice simultaneously, we first rolled one and then rolled the other. There are six possibilities for the number
showing on the first die, and then there are six possibilities for the number showing on the second. Hence,
by the Multiplication Principle, there are 6 x 6 = 36 possible outcomes in all.
PRACTICE PROBLEM SET C.F
1. 0 A box of anagram letters contains four S's, five T's, ten A's and four R's.
a) In how many ways may we select and arrange letters from the box to spell the word STAR?
b) In how many ways may we select and arrange letters from the box to spell the word START?
2. The library shelves currently contain six novels by Charles Dickens, three by Robert Louis Stevenson,
and four by Mark Twain. If you wish to read one book by each of these authors, in how many
different ways could you make the selection?
We are ready to consider Problem C.2. Try it again if you haven't already solved it.
SOLUTION TO PROBLEM C.2
There are eight possibilities (2,3,4,5,6,7,8, or 9) for the first digit, two possibilities (0 or 1) for the
second digit, and nine possibilities (1,2,3,4,5,6,7,8, or 9) for the third digit. By the Multiplication
Principle, we should expect 8-2-9 = 144 possible area codes. However, we must be careful. Rule (d) of the
problem restricts our freedom of choice for the third digit; specifically, for some choices of the second
digit, only eight choices (2,3,4,5,6,7,8, or 9) are possible for the third digit. We cannot use the
Multiplication Principle directly; instead, we proceed in either of two ways.
We can divide possible area codes into disjoint sets - those with middle digit 0 and those with
middle digit 1. By the Multiplication Principle, there are 81-9 = 72 possibilities in the first class and 8-1-8
= 64 in the second. By additivity, there are 72 + 64 = 136 area codes possible.
As another approach, of the 144 possibilities originally considered, only those which end in 11
must be eliminated. As there are eight such numbers (211,311,...,911), we have (l44 - 8 = 136) area codes.
> PERMUTATIONS
An important application of the Multiplication Principle occurs when we have a collection of
objects and we wish to order them in some way. For example, we might want to assign post positions in a
horse race, or we might wish to select Ms. America, 1^^ runner up, 2"^ runner up, 3*^^ runner up and 4^
runner up from a group of five finalists.
Given n objects that we wish to order, there are n ways in which to select the first. Once this
selection is made, regardless of which object was chosen, there remain n - 1 possible choices for the
second. After first and second are chosen, there remain n - 2 possible choices for the third. And so on. By

346
APPENDIX C
the time only two objects remain, there are two choices for next to last, and that leaves only one choice for
last. Thus, by successive applications of the Multiplication Principle, there are n{n - 1)(« - 2)...3 -ll
possible ways in which the n objects can be ordered. We introduce the symbol ! (read "factorial") to help
abbreviate our notation. Thus, for any positive integer n, «! = n{n - \){n - 2)...3 •21.
For example,
5! = 54.3-21 = 120;
6!=6-54-3-2-l=6(5!) = 720.
For reasons which will become apparent later, we define 0! to be 1. In general, «!=«•(«- 1)!
PRACTICE PROBLEM SET C.G
Compute each of the following:
8! 81
a) 7!
b) 8! c)
6!
d)
6!2!
2. Compute each of the following:
9! , 9!
a) 9! b)
0!
9!
c) - d) —
6! 6!3!
Often, it is not necessary to order all objects in a set, but rather only some of them. For example,
we may wish to select a president, vice president, treasurer, and secretary from among the membership of
some organization. We are interested not only in selecting four individuals, but we must order them too -
i.e., which one should be president, etc.
An arrangement or ordering of r objects is called a permutation. If we wish to arrange r out of n
objects, then the number of ways in which this can be done is referred to as the number of permutations of
n objects taken r at a time, and this number is denoted by „ P^. or P(n, r). By the Multiplication Principle,
we have the following formula for computing „ P^. .
Theorem C.4: For any nonnegative integers r,n,r <n, „P^ = n{n - 1)...(« - r + 1) =
n\
{n-r)\
(C.4)
That is.
The number of ways to select r objects in a specific order
out of n is
p
n^ r
The first object may be chosen in n ways; the second, in
n -1 ways; the third, in « - 2 ways,...
The r '^ object in «-(/•-]) = n-r + \ ways
For example.
The number of ways to select a president, vice
president, and secretary from among twenty people
20/^3
The president may be chosen in 20 ways; the vice
president, in 19 ways; the secretary, in 18 ways.
(Note the third person above can be chosen in
20-(3-l) = 20-2 = 18 ways.
n{n-\){n-l)...{n-r + \)
Therefore, by the Multiplication Principle
I 20^3 = 201918

PROBABILITY
347
^P, = n(n-\)in-2)...in-r + \)
n{n-\)...in-r + \)in-r)...2\
(«-/•)...2 1
Note that
20/^3 =201918
2019181716...21
(n - ry.
That is,
17 16...2 1
20!
17!
(n - r)\
For r = «, formula (C.4) becomes P =— - n\ which is consistent with our definition of 01; i.e., we can
select an ordered set of n objects from n given objects in n\ ways.
PRACTICE PROBLEM SET C.H
1. 0 Compute each of the following:
^) 10^3
^^ ^)l0^5 ^) 10^10
^)io^. ")io^o
2. Compute each of the following:
^^8^3
^)8^4 '^8^5
^)RP^ ^)rP()
We are now ready to consider Problem C.3. If you have not already solved it, try it again now.
SOLUTION OF PROBLEM C.3
Part a) of the problem asks in how many ways may we select three runners, in a given order, from
among the eighteen. This is just the number of permutations of eighteen objects taken three at a time.
18!
Thus, the medals may be awarded in ,oA = —=18 17-16 = 4896 ways. That is, there are eighteen
18 3 j3,
possibilities for the recipient of the gold medal; once first place is determined, there remain seventeen
possibilities for the recipient of the silver medal, and so on.
In part b) of the problem, we must determine how many of the 4896 possible outcomes are
favorable for the event in question. There are still eighteen possibilities for the winner of the race; but,
once the winner is determined, there are only two possibilities (the winner's teammates) for the second
place finisher. Once first and second are determined, there remains only one possibility for third. Hence,
by the Multiplication Principle, only 18'21 = 36 of the possible outcomes are favorable.
Thus, the probability that one team will sweep the event is 36/4896 = 1/136.
PRACTICE PROBLEM SET C.I
1. 0 A box of anagram letters contains seven B's, ten O's, four K's, twelve E's, six P's, and eight R's. In
how many ways may we select and arrange letters from the box to spell the word BOOKKEEPER?
2. A popular betting event at some race tracks is the trifecta, for which bettors must try to select the first
three horses in a race in their exact order of finish. In a race in which eight horses are entered, how
many different trifecta tickets are possible?
3. 0 Five slips of paper, with the numbers 1,2,3,4, and 5 respectively, are placed in a hat. A four-digit
number is formed by randomly selecting one of the slips for the first digit, then selecting one of the

348 APPENDIX C
remaining slips for the second digit, and so on. What is the probabihty that the resuhing number will
exceed 2510?
> COMBINATIONS
So far, we have considered the number of ways in which we can select an ordered set of r objects
from a given set of n objects. Sometimes, we merely wish to select r objects from the given n objects,
without ordering them. For example, we want to select a committee of r people. In this case, it does not
matter who is first, who is second, and so on; all that does matter is who the r people selected are.
The number of ways in which we can select r objects from a given set of « objects, without regard
to order, is called the number of combinations of n objects taken r at a time. It is denoted by c , C(«, r)
or . (Although the notation c more closely parallels our notation for the number of permutations.
the notation is more widely used in applications, and so we use it in the remainder of this chapter.)
To compute this number, observe that, in selecting an ordered set of r objects, we could first just
select r objects and then order them. By the Multiplication Principle, the number of ways in which we may
select an ordered set of r objects from n given objects is equal to the number of ways in which we may
select r objects (without order) multiplied by the number of ways in which a selected set of r objects may
be ordered. In other words.
Since ^ P^ and ^ P^ are known, we get
{n-r)\
or, equivalently, we have the following theorem.
Theorem C.5: For any nonnegative integers r,n,r < n ,
fj («-/•)!/•!
For example, to select a president, vice president, and secretary from among 20 people, we could
first select the three people (without regard to order) and then decide which of the three should be
president, which should be vice president and which should be secretary.
The selection of three people can be done in ways.
And the ordering of the three selected people can be done in 3 P3 =3! ways.
Since, as we saw above, the president, vice president and secretary can be selected in 20 A ways,
we have
20^3 = (^3^3 ^3;
20! r20^
Therefore,
17! 13
20] _ 20!
3 J 17!3!'

PROBABILITY 349
Note that 1 = = —-is again consistent with our definition of 0!; there is only one way to
\nj n\0\
choose n objects from a set of « objects - take them all.
Note also that
n-rj (^fj - {n - r)).{n - r)\ r\{n - r)\ {n-r).r\ \r
This is not surprising, as, whenever we select r objects from a given set of n objects, we simultaneously
distinguish n-r objects - i.e., those objects not selected.
PRACTICE PROBLEM SET C.J
1. 0 Compute each of the following:
■'(':] <"] =>(::] <:] «>(':] <';
2. Compute each of the following:
^^W '\s) ^HsJ 'HvJ '\o
PROBLEM C.3b REVISITED
Let us reconsider Problem C.3b. Since the order of the first three finishers does not matter, but
only whether or not they belong to the same team does, we may attack the problem using combinations.
/^18 1 18!
There are = —-ways of selecting three medal winners (disregarding order) from the eighteen
\^J 15!3!
contestants. Of these, only six have all three winners on the same team (i.e., all three are on team A, or all
three are on team B, etc.). Thus, the probability that all three winners are on the same team is
6 _ 6 _6-3!15!_ 6-6 _ 1
"iS^ J8^ 18! 181716 136
^3j 3!15!
(Note that this is the same answer we obtained using permutations.)
As in Problem C.3b, it is sometimes (but not always) the case that a probability problem may be
attacked using either permutations or combinations, depending on the point of view we adopt. If the
problem may be viewed in such a way that order does not matter, we may use combinations; if we can also
consider the problem so that order is relevant, then we may also use permutations. Whenever this choice of
approaches exists, it is important to be consistent in counting total possible outcomes and favorable
outcomes. We consider order in counting favorable outcomes if and only if we consider order in counting
all outcomes.
PRACTICE PROBLEM SET C.K
1. 0 A box of anagram letters contains seven B's, ten O's, four K's, twelve E's, six P's, and eight R's.
Letters are selected from the box and arranged to spell the word BOOKKEEPER. These letters are then
returned to the box and thoroughly mixed. If we again select letters to spell the word BOOKKEEPER,
what is the probability that we will choose exactly the same letters as we did before (although not
necessarily in the same positions)?
2. a) Ten basketball players decided to play a game. In how many ways could they possibly form two
teams of five players each?

350 APPENDIX C
b) Twelve basketball players decided to play a game. In how many ways could they possibly fomi
two teams so that each team has at least five players and so that every player is on one of the two
teams?
> THE BINOMIAL THEOREM
An important application of combinations is the Binomial Theorem, which deals with the
expansion of (x+y)". When several polynomials are multiplied, each term of each polynomial is muUiplied
by each term of every other polynomial and then the results are added. Thus, in particular,
2 2 2
(x + y) = {x + y){x + y) = xx + xy + yx + yy = x + Ixy + y
and
3 3 2 2 2 2 3
{x-\- y) = (x + y){x + y){x + y) = x + x y -\- xyx + xy -\- yx + yxy -\- y x-^ y
3 2 2 3
= X -\-1x y -\- 3xy + y
Note that each term of the above products is of the form cx^y^~^ where c is a constant (called
the coefficient of the term), and n is the power to which x + y is being raised.
Note further that the coefficient of the x y term is always , the number of combinations
of n things taken r at a time, as that is the number of different ways in which r ;c 's may be chosen from n
factors.
2 3
For example, the x j^ term in {x + y) comes from choosing two x 's and one y from the three
factors {x-\- y), {x + y), and {x + y). This may be done in =3 ways. I.e.,
2
{x + y) {x + y) {x + y) x y
{x + y) {x + y) {x + y) xyx
2
{x + y) {x + y) {x + y) yx .
Similarly,
2 2 2
xy + yxy ■\- y x = 3xy ;
3 3
whereas the x and y terms may each be obtained in only one way:
{x + y\x + y\x + y) x^
3
(x + y){x + y\x \y) y
This same result holds true in general; that is, we have the following theorem:
Theorem C.6 (The Binomial Theorem): For any positive integer n,
(We sometimes abbreviate the right hand side of this equation by writing
which is read "the sum, as i goes from 0 to « of, UV^~^ .")
Since the combinatorial symbols are the coefficients in the Binomial Theorem, they are often
referred to as the binomial coefficients.

PROBABILITY 351
One important consequence of Theorem C.6 which is sometimes useful in counting puzzles is
obtained by setting x = y = \ in the theorem. This gives
In other words, we have the following corollary:
Corollary C,l\ For any positive integer n
n \ \ n \ In
n
We can also obtain corollary C.l in another way. For example: by the Multiplication Principle a
student can answer a four question true false exam m l-l-l-l-l =16 ways.
We could also count in the following manner: count the number of ways in which none of the
student's answers are 'true,' + the number of ways in which one of his answers is 'true' + the number of
ways in which two of his answers are 'true' and so on. This gives
so that
lymwm:
■t.
A similar argument holds for general n,
PRACTICE PROBLEM SET C.L
1. 0 Expand each of the following:
a) (;c + >;)^ b) ^x^y)^ c) {x-y)^
2. Expand each of the following:
a) {x-y)^ b) (1-x)^ c)(2 + >;)^
3. 0 Evaluate X (" 0 ['^ ]
> INDEPENDENCE AND CONDITIONAL PROBABILITY
At times, given two experiments (or steps in a procedure), the result of one affects the outcome of
the other. Other times, this is not the case.
Definition C.3: When the outcome of each experiment (or step) of a two part experiment is not in any way
influenced by the outcome of the other, the two experiments are said to be independent.
For example, given a hat containing ten slips of paper numbered one to ten, consider the following
two step procedure.
Step one: Choose a number from the hat. Do not replace it.
Step two: Choose a second number from the hat.
Clearly the result of step one affects the outcome of step two. That is, if, for example, 6 is chosen
in step one, then 6 cannot possibly be chosen in step two. Thus experiment two is dependent on experiment
one.

352 APPENDIX C
If, on the other hand, the shp of paper chosen in step one is replaced in the hat before step two is
performed, then the outcome of step two will in no way be influenced by the result of step one, and so the
two experiments will be independent.
Just as we may speak of independent or dependent experiments, we may also speak of two events
associated with the same experiment as being dependent or independent.
Definition C.4: Two events, are independent if the occurrence of one of them does not alter the probability
that the other will occur.
For example, if we view the two step procedure described above (without replacing the slip chosen
in step one) as a single experiment, and if E\ is the event that 6 is chosen in the first step, then pr(£'i) =
1/10, as there are ten equally likely possible outcomes, only one of which is favorable. Similarly, if £"2 is
the event that 6 is chosen in the second step, and E^ is the event that 7 is chosen in the second step, then,
before the experiment is conducted, pr(£'2) = priE^) = 1/10. (Again each of the ten numbers 1,2,3,..., 10 has
the same chance as each of the others of being selected in step two.) However, once Ei has occurred, the
set of possible outcomes for step two is no longer the set often equally likely outcomes that we used in step
one. The outcome set of step two is now {1,2,3,4,5,7,8,9,10}, consisting of nine equiprobable cases;
whence pr(£'3) = 1/9, and pr(£'2) = 0 (since E2 cannot possibly occur). Thus the supposition that E\ will
occur alters the probability of E2 occurring and also changes the probability of E^ occurring. We say that
E\ and Ej are dependent events, as are the events Ei and E^.
On the other hand, if the experiment is conducted with replacement (i.e., the slip chosen in step
one is replaced in the hat before step two is performed), the outcome set of step two is exactly the same as
the outcome set of step one, and so pr(£'2) = pr(£'3) = 1/10, even if we suppose that Ei occurs. Hence, in
this case, E\ and E2 are independent events, as are E\ and£"3.
In general, in performing an experiment, the probability that a particular event E occurs may be
altered if we acquire knowledge (or if we suppose) that some other event F does occur.
Definition C.5: The new probability described above is called the conditional probability of E given F and
is denoted by pr(E\F).
Thus, in the example above (without replacing the slip), pr(£'2| E\) = 0
andpr(£:3|£:,) = l/9.
Consider another example. In rolling two dice, the probability of getting a total of eleven is, as we
saw in Problem C.la, 2/36 = 1/18.
Suppose, however, that we have already rolled the dice but do not yet know the full result, as the
white die fell off the table. If the die that is still on the table shows a 1, we would no longer think that the
probability that our total is eleven is 1/18, as, no matter what the white die shows, our total cannot possibly
be eleven. (The new outcome set is {(1,1),(1,2), (1,3),(1,4),(1,5),(1,6)}.) That is, if E^ is the event that the
total is eleven, and E2 is the event that the red die shows 1, then pr(£'i| E2) = 0.
Similarly, if the die remaining on the table shows a 5, then we would say that the probability that
our total is eleven is 1/6, as exactly one of the six equally likely possible outcomes {(5,1), (5,2), (5,3),
(5,4), (5,5) and (5,6)} yields a total of eleven. Thus, pr(£'i| £3)= 1/6, where E^ is the event that the die
remaining on the table shows a 5.
The technique used for computing pr(£'i| E^) in the example above may be used to derive a general
formula for computing pr(£'| F). Knowing (or supposing) that F occurs alters the possible outcomes of the
experiment that need be considered. Outcomes for which F does not occur are no longer possible, and,
hence, may be discarded. The outcomes that remain are exactly those which are favorable for F. Of these,
only outcomes which are also favorable for E result in E occurring. Hence, in the equiprobable case, we
have the following (compare the formula with the computations in the examples above):
Theorem C. 7: pr(£'| F) = elf, where / is the number of possible (equally likely) outcomes favorable for F,
and e is the number of these which are favorable for E - i.e., the number of outcomes which are favorable
for both E and F.
If the original experiment has n possible outcomes, then pr(F) would be fin . Furthermore, if
Er\F represents the event that both E and F occur, then pr(£'n F) = ein . Hence, we obtain pr(£'r\F)/pr(F)
= ieln)l{fln) = elf= pT(E\ F).
More generally, we have the following theorem:

PROBABILITY 353
Theorem C.8: For any events E and F associated with the same experiment
pr(£| F) = pTiEr^F)/pY{F) (C.5)
or, equivalently,
pr(£nF) = pr(F) • pr(E\F) (C.6)
(Note that formula (C.5) makes no sense if pr(F) = 0 , as we cannot divide by 0.)
Reconsidering the example of the slips in the hat, without replacement,
pr(£3n £,) = pr(£,)pr(£3| ^i) = (1/10) (1/9) = 1/90
and, with replacement,
priE^n E{) - pr(£i)pr(£3l ^i) = (1/10)(1/10) = 1/100
PRACTICE PROBLEM SET CM
1. 0 Two honest dice are rolled. Find the conditional probability that the resulting sum is seven, given
that neither die showed a 3.
2. Two honest dice are rolled, one red and one white. Find the conditional probability that the resulting
sum is seven, given that the number showing on the red die is odd.
The definition of independence may be rephrased in terms of conditional probability: E and F are
independent if the supposition that F occurs does not alter the probability that E will occur, i.e., if pr(£'|F) =
pr(£). Substituting this into (C.6), we obtain the following:
Theorem C.9: If E and F are independent events associated with the same experiment, then
pr(£nF) -pr(£) • pr(F). (C.7)
Thus, in the example above, with replacement,
pr(£3|^i) = pr(^3) = l/10
and
pr(£3n £:,) = pr(£3) • pr(^i) = (1/10)(1/10) = 1/100,
as we already discovered.
Equation (C.7) is frequently taken as the definition of independence. Note that the equation is
symmetric with respect to E and F. I.e.,
pr(£'n F) = pr(£) • pr(F) = pr(F) • pr(£) = pr(Fn £).
PRACTICE PROBLEM SET C.N
1. 0 Roll an honest die. Let £1, £"2, £3, and £4, respectively be the events: the result is even, the result is
greater than three, the result is odd, the result is greater than four.
Which of the following pairs of events are independent?
a) E\ and Ej b) E\ and £3 c) E\ and E4 d) £"2 and £4
2. Select a card at random from an ordinary 52 card deck of playing cards. Let £1, £2, £3, and £4
respectively be the events: the card is a picture card, the card is a king, the card is black, the card is a
diamond.
Which of the following pairs of events are independent?
a) £1 and £2 b) E\ and £3 c) E\ and £4 d) £2 and £4 e) £3 and £4.

354 APPENDIX C
We are now ready to consider Problem C.4a. If you have not already solved it, try it again now.
SOLUTION OF PROBLEM C.4a - THE GAME OF CRAPS
We begin by listing all the ways in which the shooter can win.
Ei: He rolls 7 or 11.
Ej. He rolls 4 and then rolls another 4 before rolling 7.
E^: He rolls 5 and then rolls another 5 before rolling 7.
E4: He rolls 6 and then rolls another 6 before rolling 7.
Es'. He rolls 8 and then rolls another 8 before rolling 7.
E^: He rolls 9 and then rolls another 9 before rolling 7.
Ey. He rolls 10 and then rolls another 10 before rolling 7.
Since these are pairwise disjoint, the probability that the shooter will win is just the sum of the
probabilities of the above events:
pr(shooter wins) = pr(£'i) + pr(£'2) + ... + pT{Ej).
We know from Problem C.l that pr(£i) = 8/36 = 2/9 .
In order for E2 to happen, two independent events must both occur - the shooter must roll 4 on his
first roll (call this event E2*) and, thereafter, the shooter must roll 4 before he rolls 7 (call this event £"2**).
(These events are clearly independent, as the first roll of the dice in no way affects later rolls.) Thus, E2 =
E2*r\E2**, and so, by formula (C.6), pT{E2) = pr{E2*) • pr{E2**).
pv{E2*) is easily seen to be 3/36 = 1/12.
To compute pr(£'2**), we may restrict our attention to the cases that either 4 or 7 occurs (as all
other cases are irrelevant to £"2, since any other roll would be disregarded and the roll would be redone).
That is, pr(£'2**) is the conditional probability that 4 occurs, given that either 4 or 7 has occurred.
Therefore, the relevant outcome set is {(1,3),(2,2),(3,1),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}, and, since three
of these give a total of four,
pr(£2**) = 3/9 = 1/3
and
pr{E2) = pr(E2*) • pr(£2**) = (l/12)(l/3) = 1/36 .
In a similar manner, pr(£'3) is equal to the product of the probability that the shooter rolls a five
and the conditional probability that a five occurs given that either a five or a seven has occurred. That is,
priE^) = (4/36)(4/10) = 2/45 .
The remaining probabilities may be computed analogously:
pr(£4) = (5/36)(5/ll) = 25/396
pr(£5) = (5/36)(5/ll) = 25/396
pr(£:6) = (4/36)(4/10) = 2/45
Py{E-j) = (3/36)(3/9) = 1/36 .
Hence, the probability that the shooter wins is
2/9 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36 = 244/495 (slightly less than 1/2).
We can indicate the various possibilities and their associated probabilities in a tree diagram
(Figured):

PROBABILITY
355
loses: pr= 1/36
loses: pr= 1/18
wins: pr= 1/36
loses: pr= 1/18
wins: pr= 2/45
loses: pr= 1/15
wins: pr= 25/396
loses: pr= 5/66
wins: pr= 1/6
wins: pr= 25/396
loses: pr= 5/66
wins: pr = 2/45
loses: pr= 1/15
wins: pr = 1/36
loses: pr= 1/18
wins: pr= 1/18
loses: pr = 1/36
Totals:
shooter loses: pr = 251/495
shooter wins: pr = 244/495
FIGURE C.l
Note that we read "and" along each of the tree branches and that Theorem C.9 applies, so that we
multiply probabilities along the branches from start to finish to obtain the probability of the total branch.

356 APPENDIX C
> REPEATED EXPERIMENTS
In discussing Problem C. 1, we saw that the act of roUing two dice could be viewed as if we first
rolled one (red) die and then rolled another (white) die. In fact, we clearly could view the experiment as if
we first rolled one die, noted the result, and then rolled the same die again. Thus, the experiment of rolling
two dice is equivalent to a double performance of the experiment of rolling one die.
In general, any experiment may be repeated any number of times. In this section, we are
interested in how the probability of events associated with repeated performances of an experiment can be
obtained from the probabilities associated with the single experiment.
In the solution of Problem C.4a, we observed that the total obtained when we roll a pair of dice in
no way affects the total obtained when we roll that same pair of dice a second time. I.e., the two
performances of the experiment are independent. This is true in general. That is, in any sequence of
repetitions of the exact same experiment, each performance of the experiment is independent of the others.
Thus, in particular, if Ei and E2 are two events associated with an experiment, and if the experiment is
performed twice, then the event that Ei occurs during the first performance of the experiment and the event
that E2 occurs during the second performance of the experiment are independent. Hence, by formula (C.6),
the probability that E\ occurs during the first performance of the experiment and that E2 occurs during the
second performance is pr(£'i) • pr(£'2).^^^
This result can be generalized to any number of repetitions of the same or even different
experiments, provided that the experiments are independent. Specifically, we have the following theorem.
Theorem C.IO: IfEiEj..., Ey^ are events respectively associated with k independent experiments G1G2...,
Gk, then the probability that E\ will occur when G\ is performed and that E2 will occur when G2 is
performed, and so on, is pr(£'i)pr(£'2)...pr(£'k), where pY(E{) is the probability that Ei occurs when 0^ is
performed.
For example, suppose a manufacturer has three machines producing electrical switches. If the
probability that the first machine produces a defective switch is 1/100, the probability that the second
machine produces a defective switch is 1/150, and the probability that the third machine produces a
defective item is 1/400, then, if one item is sampled from each machine, the probability that all three
sampled items will be defective is (l/100)(l/150)(l/400) = 1/6000000. Similarly, the probability that none
of the three sampled items are defective is (99/100)( 149/150)(399/400) = 5885649/6000000.
PRACTICE PROBLEM SET CO
1. 0 if a red, a white, and a green die are rolled, what is the probability that the red die will show an odd
number, the white die will show a three, and the green die will show a number greater than two?
2. Stu Dent is taking three courses this semester. He estimates that the probability of his getting a C or
better in Math is .80, in English is .65, and in History is .90. Assuming that his grades in the three
courses are independent of each other, what is the probability that Stu receives a C or better in all three
courses?
We are now ready to consider Problem C.5. If you have not already solved it, try it again now.
^^^: This result also follows easily from the Muhiplication Principle. If the experiment has n possible
outcomes of which/i are favorable to Ei and/2 are favorable to £"2, then, by the Multiplication Principle,
there are n possible outcomes when the experiment is performed twice. Of these, /i / are favorable to
the event E (that E\ occurs during the first performance of the experiment and E2 occurs during the
second). Hence
pr(£) =f,f2/n^ = mn){f2ln) = pr(£0 • v^Ei)

PROBABILITY
SOLUTION TO PROBLEM C.5
357
Here, we are not dealing with an equally likely outcome model of probability (e.g., each depth
charge is more likely to sink a submarine than it is to miss it). However, we are given all the relevant
underlying probabilities, and so we can still proceed, using the theorems of the chapter. Since the captain
fires the depth charges in some sequence (i.e., first one blows, then a second, and so on), he is essentially
conducting four performances of the experiment of setting off a single depth charge. If the first charge
sinks the submarine, then we need not consider what would have happened when the other three charges
exploded. Similarly, if the first charge damages the ship and the second charge either scores a direct hit or
damages it again (and hence sinks it), then the third and fourth charges do not matter. And so on. We draw
a tree diagram (Figure C.2) to indicate all the various possible outcomes.
, Sink
Miss
Miss
Miss
• 1/2
1/8
. 1/16
1/32
• 1/64
• 1/128
1/256
1/8
1/32
■ 1/64
■ 1/128
•1/256
• 1/32
1/128
1/256
• 1/128
Sunk
1/256
1/256
1/256
■1/256
1/256
Tlor
Sunk
Totals: 251/256 5/256
FIGURE C.2

358 APPENDIX C
In the diagram, the probability with which each case occurs is indicated at the right. It is
computed by using Theorem CIO and the assumption that the probability of success of a subsequent depth
charge is not affected by the knowledge that a previous charge did not sink the ship. For example, the
probability of the case that the first charge causes damage (probability = 1/4), the second charge misses
(probability = 1/4) and the third charge is a direct hit (probability = 1/2) is (l/4)(l/4)(l/2) = 1/32. If we now
add up the probabilities of all the branches on which the submarine eventually sinks, we obtain the desired
probability. Thus, the probability that Captain Dimwitty will sink the submarine using no more than four
depth charges is
1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/8 + 1/32 + 1/64 + 1/128 + 1/256 + 1/32 + 1/128 +
1/256+1/128
= 1/2 + 2/8 + 1/16 + 3/32 + 2/64 + 4/128 + 3/256
= (128 + 64 + 16 + 24 + 8 + 8 + 3)/256 = 251/256.
(Actually, there is a simpler method of solving Problem C.5, to which we will return shortly.)
So far we have considered the question, "in two performances of an experiment, what is the
probability that E\ occurs the first time and E2 occurs the second?" We emphasize that this is not the same
as the question, "in two performances of an experiment, what is the probability that E\ and E2 both occur?"
The latter question does not consider when Ei and E2 occur, only whether or not they both occur. This
question is more difficult, especially if E^ and Ei are not disjoint. We shall not attempt to answer this
question here, but the discussion above and in the rest of this section should enable the reader to handle any
problem in which the answer to such a question is called for. Basically, the technique involved is to break
the event under consideration into disjoint events whose probabilities can be found, and then to use
additivity. The difficulty, if there is one, arises in making sure that the events into which the original event
is broken up are disjoint.
This same technique also applies to the solution of the following question: In n repetitions of an
experiment, what is the probability that the event E occurs exactly k times?
The answer may be found in using a tree diagram. We illustrate the case n = 4 (see Figure C.3),
using S (success) to denote the possibility that E occurs during a particular performance of the experiment
and F (failure) to denote the possibility that E does not occur. S occurs with probability p = pr(£) and F
occurs with probability q -\ — p .

PROBABILITY
359
case 1: p<SSSS) = p *
case 2: p<SSSF) = p ♦q
case 3: pt(S S F S) = p ^qp = p^q
case 4: pi(SSFF) = p ^c^
case 5: pf(S F S S) = pqp = p q
-«.2„2
case 6: pf(S FSF)= pqpq=pq
case 7: p<S F F S) = pq ^ = p V
case 8: p<SFFF) = p q^
case 9: p<F S S S) = qp ^ = p^q
2^ _ ^2^2
case 10: p<F S S F) = qp ^q=p^q'
case 11: pt(FSFS)=qp qp = p^q^
case 12: pt(F S F F) = qp q^ = pq^
case 13: p<F F S S) = q V = P^^
case 14: pt(F F S F) = q ^q= pq^
case 15: pt(F FFS)=q^ = pq^
case 1: pt(F F F F) = q '
FIGURE C.3
Note, for example, that there are =4 cases with three S's and one F. (There are ways
to select three slots out of four in which to put the S's.) Each of these occurs with probability

360 APPENDIX C
p q = p (1 - p). Hence, the probability of exactly three 5's (i.e., E occurs exactly three times) is
4 1 3 o. •, ,
p {\- p). Similarly,
I 4 1 4 4 4
pr{E does not occur) = \q = q = {\- p)
[4] 3 3 3
pT{E occurs exactly once) = \pq = ^pq = 4p(l - p)
I 4 1 2 2 2 2 2 2
pr{E occurs exactly twice) = \p q = 6p q =6p {\- p)
pr(E occurs exactly four times) = \p = p .
In general, if an experiment is repeated n times, there are | * | ways for E to occur exactly
k times. For each of these possible cases, E occurs during a particular performance with probability p and
does not occur with probability \- p . Hence, we get k p's and n-k (1 - p) 's for each choice. We thus
have the following theorem:
Theorem C.ll: In n repetitions of an experiment, the probability that a particular event E occurs exactly k
times is
where p is the probability that E occurs in one performance of the experiment.
For example, if we toss ten fair coins, the probability that we obtain exactly eight heads is
lOV 1 Yi 1 ^ 10! 10-9 45 45
^Kl) \2) 8!2!2^^ 2-l-2'^ 2^^ 1024*
If we are interested in the probability that E occurs at least k times, then we must add the
probability that E occurs exactly k times to the probability that E occurs exactly /: + 1 times, and so on.
Thus pr(£' occurs at least k times)
Ap (■-/') +L^, P (I-P) +...+ JP . (C.8)
(l/2)'0
Thus, if we toss ten fair coins, the probability of at least eight heads is
10^ 8 2 flO^ 9 flO^ 10 rflO^ flO^ flO
= (45 + 10+l)(l/2)^^ =56/1024 = 7/128.
PRACTICE PROBLEM SET C.P
1. 0 On a roulette wheel (see Problem 2 of Practice Problem Set C.C), the 0 and 00 are green. Half of the
remaining 36 numbers are red, and the other half are black.
If the wheel is spun four times in succession, what is the probability of each of the following events?
a) All four spins come up black.
b) At least two spins come up black.
c) More blacks than reds.

PROBABILITY 361
2. If five honest dice are tossed, what is the probabihty of each of the following events?
a) Five 6's.
b) The same number showing on all five dice.
c) Three odd numbers and two even numbers.
d) At most three 6's.
PROBLEM C.5 REVISITED
Let us now return to Problem C.5. Instead of computing the probability that Captain Dimwitty
sinks the submarine, using no more than four depth charges, we compute the probability that he does not
sink it.
The sub does not sink only if all four depth charges miss, or if three miss and the fourth only
damages the sub. The probability that all four miss is (1/4) = 1 / 256 . The probability that three miss
^4^ 3
and the fourth just causes damage is (1/4) (1/4) = 4/256. Hence, the probability that the ship is not
sunk is 1/256 + 4/256 = 5/256.
Thus, the probability that Captain Dimwitty does sink the ship is 1 - 5 / 256 = 251 / 256.
PRACTICE PROBLEM SET C.Q
1. 0 If ten fair coins are tossed, what is the probability of obtaining at least one head?
> EXPECTATION
When playing a game of chance, the probability of winning is not the only important
consideration; the potential payoff and the expected payoff are other important factors which should be
considered in deciding whether or not to play.
Potential payoff refers to the amounts of money that could possibly be won or lost. Few of us
would be willing to risk $1000 against the possibility of winning a nickel, even if our chances of winning
are very high; on the other hand, many of us would be willing to risk a nickel for the opportunity to win
$1000, even if our chances of winning are very low. We will return to this point later.
Expected payoff, on the other hand, refers to the amount we theoretically expect to win or lose if
we play the game a large number of times. For example, consider the following game: We roll a die. If
we get a two, the bank pays us $4; otherwise, we pay the bank $1. (I.e., the bank is giving us four to one
odds that we do not roll a two.) If we play this game six hundred times, then, since we expect each face of
the die to show up approximately equally often, we anticipate rolling a two one hundred times. We will
therefore expect to win $4 one hundred times and lose $1 five hundred times, for a net loss of $100. That
is, our expected payoff if we play the game six hundred times is -$100, which makes our average expected
payoff per game -$100/600 (approximately -$.17). (Equivalently, we expect to win $4 one sixth of the
time and lose $1 five sixths of the time so our expected payoff per game is (1/6) • $4 -(5/6)$ 1 = -$.17.)
This does not mean that we will lose $.17 every time we play the game - the only possible amounts that
could be won or lost at a given play are $4 or $1. Rather, it means that, if we play the game a large number
of times, we expect, in the long run, to lose money; in fact, we expect to lose an amount approximately
equal to $.17 muhiplied by the number of games we play. Obviously, if we are lucky, we could
conceivably win (or, if we are unlucky, we could lose larger amounts), but the laws of probability tell us
that it would be unwise for us to play the game in question.
Similarly, if we are offered odds of 5 to r that some event E does not occur (i.e., we win $s if E
occurs and lose $ r otherwise), and if the probability of E occurring is p , then our expected payoff is
sp - r(l - p). If this is negative, then we expect to lose money in the long run, and we say the odds are not
favorable to us. If it is positive, we expect to win money in the long run, and say the odds are favorable to

362 APPENDIX C
US. Finally, if sp - r(\ ~ p) = 0, then, in the long run, we expect to break even, and we say the odds are fair.
In this case, sp = r(l - p), and so
^ _ P
s 1 - p
Note that fair odds correspond to the theoretical definition of odds given at the beginning of the
chapter (Definition C.2).
PRACTICE PROBLEM SET C.R
1.0 If someone hands you a pair of dice and offers to bet that you do not roll a double (two like
numbers), what odds should you be offered in order for the bet to be a fair bet?
2. A roulette wheel contains each of the numbers 1 to 36 as well as 0 and 00. If you bet on a number and
it turns up, you win thirty five times the amount of your bet, i.e., odds of 35 to 1. Are these odds fair?
If not, what would fair odds be?
In general, a game of chance may have more than two possible payoffs. Suppose that the possible
outcomes of a game are grouped into k disjoint events, £^£2 ..., £'k, which occur with probabilities/? 1/72 •••,
Pk respectively O^i + /?2 + ••• + Pk = 1)- Suppose further that these events respectively require payoffs of
m\m2..., mk. (I.e., if £"1 occurs, then you win mj if mj is positive, lose mi if mi is negative, and neither win
nor lose if mj = 0.)
Definition C.6: In the situation described above, the expectation or expected value of the game is
e = mi/7,+m2/?2+--. +^"1^^.
(The rationale behind this definition is essentially the same as in the case of two payoffs. Try to supply the
details yourself)
For example, suppose we have the following arrangement: We roll one die. If we roll a one, we
win $1; if we roll a three, we win $3; for a five, we win $5; otherwise, we lose $2. Then the expected value
of the game is 1(1/6) + 3(1/6) + 5(1/6) - 2(1/2) = 1/2 = $.50, since the probability of rolling a one is
1/6, the probability of rolling a three is 1/6, the probability of rolling a five is 1/6, and the probability of
rolling something else is 1/2.
We are now ready to solve Problem C.4b. Try it now if you haven't already solved it.
SOLUTION OF PROBLEM C.4b
We distinguish five events.
E\: The shooter loses.
E2: The shooter wins by rolling 7 or 11.
Ey. The shooter wins by making his point, which is 6 or 8.
E4: The shooter wins by making his point, which is 5 or 9.
Es'. The shooter wins by making his point, which is 4 or 10.
From the solution of Problem C.4a, the probability of each of these events is easily found. (See
Figure C.4.)
Event
£1
E2
E3
E,
Es
Probability
Payoff
-1
1
1
3/2
2
FIGURE C.4

PROBABILITY 363
Hence, the expected value of the game is
2(2/36) + (3/2)(4/45) + 1(50/396) + 1(2/9) - 1(251/495)
= 85/990 ^.086.
That is, in the long run, the shooter expects to win an average of about 8.6 cents for each dollar he
bets.
PRACTICE PROBLEM SET C.S
1. 0 Select a card at random from an ordinary 52 card deck of playing cards. If you select a spade, you
win $3; if you select a red picture card, you win $5; otherwise, you lose $1. What is the expected value
of the game?
2. In a game of roulette, you place a $1 bet on red and a $1 bet on 11 (which happens to be a black cell). If
a red number comes up, you win $1. If 11 comes up, you win $35 dollars; otherwise you lose $2. What
is the expected value of the game?
We reiterate that we do not win or lose the expected value of a game each time we play. However,
if we play the game often, then we expect our net gain (or loss) to be approximately the expected value
times the number of games we play.
If the expectation of a game is positive, we expect to win in the long run, and we say that the game
is favorable; if it is negative, we expect to lose, and say that the game is unfavorable; if the expectation is
zero, we expect to break even. In the latter case, we say that the game is a fair game. That is, we have
Definition C. 7: A fair game is a game with expected value zero.
Any gambling game may be converted to a fair game by requiring one of the players to pay for the
privilege of playing. For example, if a player's expectation is +$2 and if he pays $2 for the privilege of
playing, then the game becomes fair.
PRACTICE PROBLEM SET C.T
1. 0 You toss three fair coins and win a number of dollars equal to the number of heads you obtain. What
should you pay for the privilege of playing this game in order to make it a fair game?
We emphasize that it is possible to lose at a favorable game and to win at an unfavorable game
(otherwise, no one would gamble). However, the more often a game is played, the less likely it is for this
to happen.
We are now ready to solve Problem C.6. Note that part b) of the problem is just a special case of
the problem which led Pascal and Fermat to develop the theory of probability.
Now it is your turn to try to solve the problem using the theory that has been developed in the chapter.
SOLUTION OF PROBLEM C.6
In this problem, there are several different probabilities in which we are interested. There is the
probability that A will win a point at any given roll of the die. This probability remains constant
throughout the game (successive rolls are independent) and is easily seen to be 1/3 (two of the six outcomes
are favorable).
Similarly the probability that B will win a point at any given roll of the die is 2/3.
In addition, since a fair distribution of the money in the pot will depend on the relative
probabilities that A and B will be the uhimate winner, we are interested in these probabilities. These
probabilities change as the game progresses and the score changes.

364 APPENDIX C
In part a) of the problem, we are interested in the probabiUty before the game begins that A will be
the eventual winner. Let Pa denote this probability. Define Pb similarly.
Let Sfi, and 5b denote the sums of money anted by A and B respectively. Then A's expectation is
Pa 5b-Pb 5a and B's expectation is Pb 5a-Pa 5b.
Since the game is to be a fair game Pa 5b - Pb 5a = 0, or, equivalently,
Pa 5b = Pb 5a
Since 5a + 5b = 100 and Pa + Pb = 1,
Pa (100 - 5a) = (1-Pa) 5a
100 Pa-Pa ^a = 5a -Pa5a
100 Pa = 5a
Therefore, we will know Sa if we can find Pa-
How can A be the winner? He can win by a score of 4-0, 4-1, 4-2, or 4-3.
In the first case he must win the first four points. This occurs with probability
Qr-
If he wins by a score of 4 to 1, he must win the fifth point and three out of the first four points
(losing the remaining one). This occurs with probability ~ ~
Similarly he wins by a score of 4-2 with probability —
4/^x3
6 ,n2
and he wins by a score of 4-3 with probability ~ ~" I • Thus
= 379/2187-.1733
Therefore
5a =100 Pa =$17.33
That is, A should ante $17.33 and B should put up $82.67.
For part b) of the problem, since A has two points and B has one, A needs only two points to win
and B needs three. Thus the probability that A would have won had the game continued is the probability
that A would have gotten two more points before B gained three more. Again this could happen in several
ways:
A wins the next two points - probability = —
A wins the third point and one of the next two - probability = ~ ~
2 / \ 2
A wins the fourth point and one of the next three - probability = | —
Thus, had the game continued, the probability that A would have won is

PROBABILITY 365
Therefore, the probability that B would have won is 16/27 ~ .5926. Hence the pot should be divided in the
proportion 4074 to 5926. That is, A should get $40.74 and B should get $59.26. (Note that A wins
$23.41.)
> ENTERTAINMENT VALUE AND POTENTIAL PAYOFF
In general, most games of chance operated by gambling casinos or racetracks are slightly
unfavorable to the player -i.e., the player's expectation is slightly negative. This is necessary in order for
the gambling establishment to stay in business. Under these circumstances, why do many intelligent people
choose to gamble anyway? The answer seems to involve at least two factors.
Firstly, participation in a game of chance is a form of entertainment. Just as one might be willing
to spend money to attend a show, concert, movie, sporting event, etc., or to eat at a fancy restaurant, one
might also be willing to spend a certain amount of money simply for the enjoyment of gambling. Since, in
the long run, a player should expect to lose an average of "e" for each time he plays a game having
(negative) expectation, -e, we say that e is the entertainment value of the game. That is, the player is
essentially willing to invest e per game, just for the fun of playing.
The second, and probably more important, factor which influences people to gamble is the
possibility, no matter how unlikely, of winning a big payoff. The person who invests one dollar per week
in lottery tickets usually figures that losing a dollar a week is not going to have a great effect on his or her
standard of living; but, imagine what would happen to his (her) standard of living if he (she) should, by
some miracle, win one million dollars.
Some of us apply the same principle, on a smaller scale, to sweepstakes sponsored by magazines
and mail order houses. In this case, the only investment required of the player is a postage stamp (at the
time of this writing $.13). However, chances of winning the grand prize are usually so small that even a
$.13 investment is not warranted unless the potential payoff is sufficiently large. To our mind, prizes of
$5000 or even $10000 are not sufficiently tempting^^ to be worth the investment of $.13, chances of
winning being as small as they are. (This is not to say that we would not be thrilled to win even $1000.)
However, we find prizes of $100,000 or more sufficiently enticing to justify the investment of $.13, even
though we know in our heads and hearts that we will not win.
> THE CHAPTER IN RETROSPECT
Our object in this chapter has been to introduce you to the notions of probability, expectation,
odds, etc., not only so that you can gamble more intelligently but also since these words are becoming more
and more part of our daily vocabulary.
Although we have basically considered probability as defmed in the case of equally likely
outcomes, many of the results hold in general. Remember though, that, in the general case, care is needed
in deciding on how the outcomes should be weighted.
It is extremely important to be able to count cases properly, and so we have considered several
types of counting techniques. Make sure you understand thoroughly when each type can be used. Try now
to apply these ideas in the exercises which follow.
^^^: We never could understand the thinking of a bank robber in movies which have him steal
$io,ooo or even $50,000 vnth the expectation of escaping to South America and being able to
live the rest of his life like a king.

366
APPENDIX C
EXERCISES
Introductory Problems
C.i [h| @ a) Remove nine cards (AS, 2S, 3S, ...,
9S)
from a deck of playing cards.
Place them in a number of piles so that the
sum of the face values of the cards in each pile is
the same. Disregarding the order of the piles, in
how many ways can this be done?
b) What if ten cards (AS,..., loS) are used?
C.2. |h| § How many triangles (of all sizes) are
there in Figure C.5?
FIGURE C.5
C.3. § Winnie Gamble loves to play the slot
machines. She has nineteen silver dollars and one
slug (which will be rejected by the machine) in her
purse.
a) If Winnie randomly selects one coin from her
purse, what is the probability that it will be the
slug?
b) What is the probability it will be a good
coin?
c) If Winnie selects four coins from her purse,
what is the probability that at least one of them
will be the slug?
C.4.§ In every four hundred year period, the
thirteenth of a month falls on a Friday 688 times
and the first day of a century falls once on
Monday, once on Wednesday, once on Friday, and
once on Saturday (see Exercise 4.36).
a) If a year and a month are selected at
random, what is the probability that the
thirteenth of that month falls on a Friday in the
selected year?
b) What is the probability that the first day of
a century falls on a Sunday?
c) What is the probability that the first day of
a century falls on a Monday?
C.5. § Jack and Jill each selects two distinct
odd numbers between three and seventeen
inclusive. They then simultaneously throw three
dice each. Whoever first rolls his or her dice and
obtains a sum equal to one of his or her own
selected numbers is the winner, unless his or her
opponent is also successful, in which case the
game starts over again.
Jack and Jill have selected distinct pairs of
odd numbers so that they have equal chances of
winning.
If Jack has selected seven as one of his
numbers, what numbers has Jill selected, and
what is Jack's other number?
Counting, Using The Multiplication
Principle, Permutations, Combinations
And Additivity
C.6. § Don and Dinah went to a diner for dinner.
The menu is shown in Figure C.6.
Appetizer
Fruit juice
Soup du jour
Melon
Entree
Filet mignon
Breaded veal cutlets
Fried flounder
Beef stew
Dessert
Chocolate mousse
Ice cream
Sherbet
Cheese Cake
Fresh Fruit
FIGURE C.6
a) How many different complete dinners
(including one appetizer, one entree, and one
dessert) are available?
b) Dinah could not make up her mind about
what to order, so she told the waiter to surprise
her. Assuming that the waiter randomly selected
one dish in each category, what is the probability
that Dinah had a liquid appetizer, a meat entree,
and either ice cream or sherbet for dessert?
C.7 |h| § The repair lady for the telephone
company has been given a list of five phone
numbers which are in need of repair (see Figure
C.7).
Phone Number
527 - 5263
468 - 4486
643 - 3546
266 - 7269
267 - 8666
Signature
FIGURE C.7
After each phone is repaired, she must
obtain the signature of each of the people in
whose homes the phones are located. Assuming
that no two names are the same, what is the
probability that, when the signatures are

PROBABILITY
obtained, the last names of the people will be in
alphabetical order?
C.8. @ There was one other number that the
repair lady was supposed to fix, but she forgot to
bring the number with her. However, she does
recall that the last four digits of the number were
all distinct.
What is the probability that the last four digits
of the number begin with a 2 and contain a i?
C.9. @ In how many different ways can a cube be
placed in a box of the correct size?
C.io. |§ § Given six colors, in how many ways
can the faces of a cube be colored so that each face
is a different color? (Two colorings are
considered to be the same if the resulting cubes
can be so placed so that the same colors face the
same directions.)
C.ii. @ Mrs. Cooke is having a big dinner party
and has decided to start the dinner with stone
soup. Her recipe calls for two large stones, one
cup each of any five vegetables, diced, and one
quarter pound of meat, for each two gallons of
water. (Note, you must never mix two varieties of
meat in the same pot of stone soup!)
a) If Mrs. Cooke has already placed two
stones and the meat in two gallons of boiling
water, and if she has twelve varieties of vegetables
to choose from, in how many ways could she
choose which five vegetables to add to the soup?
b) After the soup was finished, Mrs. Cooke
cooked up another potfull. (She could not have
doubled the recipe in the first place, because she
did not own a pot that would be large enough.)
She had six stones remaining to choose from, as
well as seven varieties of vegetable and three
different kinds of meat.
In how many different ways could she make
the second pot of soup?
C.12. § Sandra, Jessie, and Veronica bowl
together every Thursday night. This Thursday,
however, each, unbeknownst to the others,
decided that she would rather attend their college
basketball game. Not wishing to offend her
friends, each of the girls told the others that she
was sick in bed and could not bowl. Fortunately,
they were all sitting in different locations in the
basketball arena, so they did not notice each
other. However, at the exact same moment, all
three decided to go to the powder room. If there
are three powder rooms in the building, what is
367
the probability that none of the girls will find out
that anyone else lied?
C.13. |h| @ At the Nationalist Party's fundraising
dinner, there were nine tables arranged in a
square. There are three red, three white and three
blue tablecloths.
In how many ways can they be placed on the
tables so that no two tablecloths of the same color
are in the same row or the same column?
C.14. |h| @ John, Joan and Jack Smith, John,
Joan, and Jack Brown, and John, Joan and Jack
Jones all had tickets to the same performance at
the opera. Their seats were in rows D, E, and F,
and were numbers 101,102, and 103 in each row.
a) In how many ways can the tickets be
distributed so that no members of the same family
have tickets with the same number or letter?
b) In how many ways can the tickets be
distributed so that not only do no two members of
the same family have tickets with the same letter
or the same number, but also the same is true of
people with the same first name.
c) Assuming that the conditions referred to in
part b are realized, what is the probability that
Jack Smith is sitting immediately to the right of
Joan Brown?
**C.i5. |h| @ The four Bridges, Manny, Brooke,
William, and Queenie, play bridge together four
evenings a week. If, at the beginning of each
week, they make a seating plan so that each
person sits in a different direction (North, East,
South, or West) each of the four evenings that
week, how many weeks could pass without their
having to repeat a complete weekly seating plan?
C.16. § Mr. and Mrs. B. A. Scout are going on a
camping trip with their baby. They figure that
they will need ten jars of baby food for the trip.
Their pantry is stocked with eight jars of baby
fruits, twelve jars of baby vegetables, and six jars
of baby meats. If they select ten jars at random,
what is the probability that

368
APPENDIX C
a) they select five jars of vegetables, three of
fruit, and two of meat?
b) all ten jars are the same (i.e., all fruit, all
vegetables, or all meat)?
c) at least one type of food is missing (i.e.,
there is no fruit, or there is no vegetable, or there
is no meat)?
**C.i7. |h| § Given the six colors and the cube of
Exercise C.io, how many colorings are possible if
more than one face may receive the same color?
E.g., the entire cube may be colored red, etc.
Remember that two colorings are considered to be
the same if the resulting cubes can be so placed
that the same colors face the same directions.
C.i8. @a) How many different three letter words
(meaningful or not) can be written using the
English alphabet?
b) (H) How many are there if the middle letter
must be a vowel and the last must be a consonant,
and if no more than one "y" is to be used in any
word? ("y" is to be considered as a consonant
unless there is no other vowel in the word.)
C.19. § § a) How many four digit numbers
may be formed using only the odd digits
(1,3,5,7,9)?
b) What is the sum of all these numbers?
c) If these numbers are arranged in numerical
order from smallest to largest, what is the 314^^
number on the list?
*C.20. |h| § a) How many numbers of no more
than four digits may be formed using only the odd
digits?
**b) What is the sum of all these numbers?
If these numbers are arranged in numerical order,
what is the 314^^ number on the list?
*C.2i. |h|§ a) How many four digits numbers
are there whose digits are distinct?
**b) What is the sum of all these numbers?
c) If these numbers re arranged in numerical
order, what is the 314^^ number on the list?
C.22. § In how many ways can three red and
four black checkers be arranged in a straight line?
(Consider checkers of the same color to be
indistinguishable from each other.)
C.23. |h| § In how many ways can four red and
four black checkers be arranged in a straight line
so that the n^^ ^ed checker (counting from the
left) is not to the left of the n^^ black checker, for
n = 1, 2, 3, and 4?
C.24. § § The elections committee was holding
an election, by secret ballot, for chairperson of the
committee. The final vote was 4 for Delia Gate
and 6 for Victor Riaz. The votes were counted one
at a time.
a) What is the probability that Delia was
ahead when the fifth vote was counted?
b) What is the probability that Delia was ever
ahead in the counting of the ballots?
C.25. |h| § Ima Gardner has enough space to
plant five fruit trees in a row on her property. She
has been given two apple, two pear, two peach,
two plum, one apricot and one cherry tree from
which she may choose any five.
How many different arrangements are
possible? (Trees of the same type are to be
considered as indistinguishable.)
C.26. |h| [60, volume 1, page 85] a) In how many
ways can you place two kings, one white and one
black, in a legal position (i.e., in nonadjacent
cells) on an 8 X 8 chessboard?
b) In how many ways can you place two rooks
(white and black) so that they do not attack each
other? (I.e., they are not in the same row or file.)
c) In how many ways can you place two
bishops (white and black) so that they do not
attack each other? (I.e., they are not on the same
diagonal.)
d) In how many ways can you place two
queens (white and black) so that they do not
attack each other? (I.e., they are not in the same
row, file, or diagonal.)
e) In how many ways can you place two
knights (white and black) so that they do not
attack each other? (See Chapter Six for a
description of the knight's move.)
C.27 |H| |A| Using the dots in a 6 x 6 array (Figure
C.8) as vertices,
a) How many squares having horizontal and
vertical sides can be drawn?
b) How many rectangles which are not
square can be constructed having horizontal and
vertical sides?
*c) How many squares can be drawn with
sides which are not horizontal or vertical?
FIGURE C.8

PROBABILITY
369
C.28. @@ How many rectangles (including
squares) having horizontal and vertical sides can
be drawn on an n x m array of dots?
C.29. |h| § A man has three dice. On one of
them, the one has been changed to a second three;
on another, the six has been rubbed out and the
face is blank.
a) Counting the blank as zero, how many
different numbers can be made by turning the
dice any way you like and placing the top faces
together. For example, the dice in Figure C.9 are
showing the number 305.
A __/ Z /• /i —V •
I r 1 [•
J J \ *
I * / * • */ * \/
FIGURE C.9
b) If the dice are rolled, what is the
probability that the top three faces show a three, a
five, and a six (regardless of order)?
C.30. § a) In how many different ways can ten
people line up in a row?
b) In how many different ways can ten people
line up in a row if two particular people, say,
Curtis and Clementine, must be next to each
other?
c) |h| How many different circular
arrangements often people are there?
d) How many circular arrangements are
there if two particular people, say, Curtis and
Clementine, must be next to each other?
C.31. § How many different necklaces can be
made from seven beads, each of a different color?
(Two necklaces are considered the same if they
can be turned or flipped over so that they appear
identical.)
C.32. |h| 0 How many different necklaces can be
made so that each contains four red, three blue
and two yellow beads?
C.33. 1^ § Given a supply of four red, four green,
and four white beads, how many different six
bead necklaces could be made?
*C.34. @ If a bridge hand (thirteen cards) is dealt
from an ordinary fifty-two card deck of plajdng
cards, find an expression for the probability that
a) all thirteen cards are of the same suit
(clubs, diamonds, hearts, or spades)?
b) § at least eight cards are in the same suit?
c) at least seven cards are in the same suit?
**d) § at least six cards are in the same suit?
e) exactly two different suits are represented?
f) two suits contain ten or more cards between
them, but no suit contains seven or more cards?
g) § there are four cards in one suit and
three cards in each of the others?
h) there are four cards in each of three suits
and one card in the fourth suit
i) there are no more than four cards in any
suit?
j) there are five cards in one suit and no more
than three cards in any other suit?
k) the hand contains four or more picture
cards (jacks, queens, and kings)?
1) the hand contains all four aces?
m) the hand contains no card lower than a
ten? (Consider aces to be higher than a ten.)
n) the hand contains no card higher than a
ten?
*C.35 0 A hand containing five cards is dealt
from an ordinary fifty-two card deck of plajdng
cards. What is the probability that
a) the hand is a flush? (I.e., all five cards are
in the same suit.)
b) no two cards are in the same suit?
c) § the cards form a straight? (I.e., five
cards which can be arranged, regardless of suit, to
form a consecutive sequence such as A,2,3,4,5;
2,3,4,5,6;..., 9,io,J,Q,K; or io,J,Q,KA)
d) the cards form a straight flush? (I.e., they
are in the same suit and can be arranged to form a
consecutive sequence.)
e) four cards in the hand have the same face
value. (Such a hand is referred to as four of a
kind.)
f) § the hand is a full house? (I.e., three of
the cards have the same face value, and the other
two cards have the same face value as each other.
The face value of the two cards is obviously
different from that of the three cards.)
g) § three of the cards have the same face
value, but the other two have different, distinct
face values. (Such a hand is referred to as three of
a kind.)
h) two of the cards have the same face value,
and the other cards have different distinct values?
(Referred to as two of a kind, or one pair.)
i) two of the cards have one face value, two
other cards have the same face value as each other
(different from that of the first two cards), and the

370
APPENDIX C
remaining card has a third face value? (This is
referred to as two pairs.)
j) |h| no two cards have the same face value
and the hand is not a straight nor is it a flush?
*C.36. @ @ In poker, a straight flush (see Exercise
C35) beats four of a kind, which beats a full
house, which beats a flush, which beats a straight,
which beats three of a kind, which beats two pair,
which beats one pair, which beats no pair. In a
game of draw poker, you are dealt the following
five cards: loS, JS, QS, KS, KH. You are allowed
to throw away up to four cards and to replace
them by new cards drawn from the deck.
a) If you keep only the two kings, what are
the probabilities that you will obtain a straight
flush, four of a kind, etc?
b) If you keep the four spades, what is the
probability of each type of hand?
**C.37. § @ In a game of five card stud poker,
two of your opponents are currently showing the
follovsdng hands: (You do not know what the face
down cards are.)
QQaS,3H,2D, and n,D6H,8H,9H,
Two other players have already folded (i.e., given
up), having shown KS,6D,5S, and 2H between
them.
You have 5D,5H,5C,7S,ioC.
What is the probability you hold the winning
hand? (Refer to Exercises C.35 and C.36 for the
various types of hand and their relative values.)
Note that three aces beats three kings which beats
three queens, etc.
C.38. §§ Three people participate in a grab bag.
Each brings a gift. These are then placed in a bag,
and each person selects one of the gifts, sight
unseen
a) How many different outcomes are
possible, if no one is to receive the gift that he or
she brought?
b) How many outcomes are possible if four
people participate and if no one receives the gift
that he or she brought?
c) Five people?
*d) Six people?
e) If six people participate, and if no one
receives the gift that he or she brought, what is
the probability that Ross receives the gift brought
by Sanford and that Sanford receives the gift
brought by Ross?
*C.39. |H| |A| The ace of spades, two of spades, ....,
ten of spades are all removed from a deck of
cards, shuffled, and placed face down on the table.
You turn over the first card, saying "one," turn the
next card, saying "two," and so on. What is the
probability that the ace will come up when you say
one, or that the two will come up when you say
two, etc? I.e., that you name at least one card
correctly as you turn it?
Conditional Probability And Independence
C.40. @ During a game of draw poker. Cord
Sharpe "accidentally" dropped one of his five
cards on the table. It was a deuce (two). What is
the probability that Cord holds at least one more
deuce in his hand?
C.41. @ In a game of Craps, your point is six.
What is the probability that
a) you win in no more than three additional
rolls?
b) you lose in no more than three additional
rolls?
C.42. § When Una and Zvi Dreier got married,
they decided to have exactly three children.
Assuming that each child is as likely to be a boy as
it is to be a girl and that there were no multiple
births except as indicated,
a) what is the probability that the Dreiers
have three boys?
b) what is the probability that the Dreiers
have three boys, given that the first born is a boy?
c) what is the probability that the Dreiers
have three boys, given that at least one of the
children is a boy?
d) what is the probability that the Dreiers
have exactly three boys, given that Una gave birth
to identical twins at her third delivery?
C.43. |§ § Referring to Exercise C.14, and
assuming that the conditions referred to in part b)
of the problem are realized, what is the
probability that Jack Smith is sitting immediately
to the right of Joan Brown?
C.44. § @ In a game of Blackjack (Twenty One),
using a single ordinary deck of fifty two cards, you
have a nine and a six, totaling fifteen. The dealer
has a nine showing.
a) What is the probability that you can draw
another card so that your total will not exceed
twenty one? (Picture cards count as ten; aces as
one or eleven, as you wish. If you exceed 21, you
lose and the game ends.)
b) By looking around the table and
remembering cards which have already been
played, you can account for nine picture cards.

PROBABILITY
371
two tens, all four nines (counting yours and the
dealer's), two eights, one seven, two sixes
(including yours), three fives, all four fours, three
threes, all four twos, and all four aces. In this
case, what is the probability that, if you draw
another card, your total will not exceed twenty
one?
c) The rules of the game state that the dealer
must continue to draw cards until his total is
seventeen or more, at which point he may not
draw any more cards. If his total exceeds twenty
one, he loses.
Using the information given in part b), what
is the probability that, if you stand pat at fifteen
(i.e., take no more cards) then the dealer will beat
you (i.e., he will have a total between seventeen
and twenty one, inclusive)?
*d) Should you draw or not? (In the event
that you and the dealer have the same total which
is less than twenty one, then no one wins.)
*C.45. |h| § Repeat Exercise C.44 in the case that
the dealer has a six showing rather than a nine.
(You can now account for three nines and three
sixes rather than four nines and two sixes.
Everything else is as in Exercise C.44.)
C.46. § Eight people wish to choose up sides to
play a game of football. They decide to choose by
the method often referred to as "the four same
fingers." That is, they all chant, "The four same
fingers are together," and, on the word
"together," each person extends either one or two
fingers on the right hand. If four people have
extended one finger and the other four have
extended two, then the two teams are determined;
otherwise, the process is repeated again.
Assuming that each person is just as likely to
extend one finger as two,
a) what is the probability that the method will
be successful on the first try?
b) what is the probability that Dan and Eliot
will be on the same team?
C.47. § @ A second method for choosing up two
four person teams out of eight players is as
follows: One of the eight is separated from the
rest. The remaining seven then choose as in
Exercise C.46, with the "four same fingers"
constituting one team and the designated player
then belonging to the other team with the "three
same fingers."
a) What is the probability that this method
will be successful on the first try?
b) If Dan is the designated player, what is the
probability that Dan and Eliot are on the same
team?
c) If neither Dan nor Eliot is the designated
player, what is the probability that they will both
be on the same team?
*C.48. § A third method of choosing teams from
eight players is to have two designated players
and to have the remaining players choose as in
Exercise C.46. If there are "four same fingers,"
then the two designated players are on the same
team with the "two same fingers." If there are two
groups of "three same fingers," then the
designated players flip a (fair) coin to see who is
on which team. If the choosing process comes up
five to one or six to zero, then it is repeated.
a) What is the probability that the method
will be successful on the first try
i) without requiring a coin toss?
ii) requiring a coin toss?
b) If Dan and Eliot are the two designated
players, what is the probability they end up on the
same team?
c) If Dan is one of the designated players but
Eliot is not, what is the probability that they will
end up on the same team?
d) If neither Dan nor Eliot is a designated
player, what is the probability they will end up on
the same team?
C.49. |h| § If there are twenty five people in a
room, what is the probability that at least two of
them have the same birthday? (Assume that none
have birthdays on February 29, and that all other
birthdates are equally likely.)
Repeated Trials
C.50. § A student is taking a ten question
multiple choice exam, each question offering four
choices. If the student answers the questions
completely at random, what is the probability that
a) he gets everything right?
b) he gets everything wrong?
c) he gets eight or more right?
C.51. @ Senta Dead, Cy Bull, and Mark Hitda
were sent behind enemy lines to blow up the
bridge on the Kly River. They had planted their
explosives and were about to blow up the bridge
when they discovered that the detonator was not
working properly. Their only hope of a successful
mission was to shoot the dynamite, which had
been placed in three key locations. Unfortunately,
they each had only one bullet remaining.

372
APPENDIX C
Fortunately, all were excellent markspeople. At
the distance from which they had to shoot, Cy
would hit his target 95% of the time, Senta would
not miss more than 10%, and Mark had an 80%
probability of success. They decided that they
would fire simultaneously, each at a different one
of the key dynamite placements. Assuming that
the success of each of the three is independent of
the success of each of the others,
a) what is the probability that all three hit
their targets?
b) what is the probability that at least two hit
their targets? (This would probably be enough to
bring down the whole bridge anyway.)
c) what is the probability that they at least
damage the bridge? (I.e., they have at least one
hit.)
C.52. @ Ten slips of paper, numbered o to 9
respectively, are placed in a hat. A slip is
selected, its number noted, and then it is replaced.
The procedure is repeated until a total of five
numbers have been noted.
a) What is the probability that all five
numbers are the same?
b) What is the probability that all five
numbers are different?
c) What is the probability that at least two
numbers are the same?
C.53. § Ty M. Waster spends all his free time
playing with a game of the type pictured in Figure
C.io. A metal ball is inserted through the opening
at the top of the game, and, as the ball falls, it hits
the metal nails and either falls to the right or left,
eventually landing at the bottom.
a) How many different possible routes are
there that the ball could follow on its way down?
b) For each of the six bottom slots, how many
routes end in that slot?
c) What is the probability that a ball placed at
the top of the game will land in slot number two?
12 3 4 S
FIGURE C.io
*C.54. g @ Last time I went to the theater, I
noticed that there were ten seats in the row in
which I was seated and that, although the seats
were filled one at a time, at no time were there
vacant seats between seats that were already
occupied.
In how many different ways could this occur?
C.55. § The Women's Auxiliary of the
Amalgamated Plumbers of Watertown is holding
its annual gala affair, and each of the nine
members has been asked to bake either an angel's
food cake, a devil's food cake, or a marble cake. If
each woman is just as likely to bake any one of the
three as she is to bake any of the others, what is
the probability that
a) there are five angel's food cakes, three
devil's food cakes, and one marble cake?
b) there are five or more angel's food cakes?
c) the number of angel's food cakes exceeds
the number of devil's food cakes?
Odds And Expectation
C.56. § At Upson Downs, four horses are entered
in today's feature race. The odds on Far Behind
are 5 to 1; on Dead Heat, they are 2 to 1; on
Running Last, 11 to 1. If these are fair odds (i.e.,
the track is not taking a percentage), what are the
odds on the fourth horse. Favorite Son?
C.57. § In the game of Chuckluck (also known as
Bird Cage) there are three dice in a cage. The cage
is rotated, acting as a shaker.
a) What is the probability that the sum of the
top faces of the three dice will be thirteen?
b) What is the probability that at least two of
the dice will show a two?
c) Before each roll, money can be bet on any
of the numbers 1, 2, 3, 4, 5, or 6. If you bet on a
number and if that number shows up on one of
the dice, then you win the amount of your bet; if it
shows up on two dice, you win twice the amount
of your bet; if it shows up on all three dice, you
win three times the amount of your bet.
Otherwise, you lose.
What is the expected value of this game?
(Assume one dollar bets.)
C.58. @ Raphael Loff was in need of $200, but
his paycheck was only in the amount of $175. He
therefore decided on the following plan. He held
a raffle, selling two hundred raffle tickets at $1
each, and offering his paycheck as the prize. If
Tekka Shantz bought one ticket, what was her
expectation?

PROBABILITY
373
C.59. @ Rudolph, Thomas and Neil each has
three denominations of U.S. currency in his
pocket. Rudolph has one $5 bill, one $20 bill, and
one $100 bill; Thomas has one bill each of $1,
$50, and $500; Neil has one bill each of $2, $10,
and $1000.
a) If Rudolph and Thomas each randomly
selects one bill from his pocket, who is more likely
to select the bill with the higher denomination?
b) What if Thomas and Neil each selects one
bill?
c) What about Neil and Rudolph?
d) Each of the three selects one bill, at
random, from his pocket, and they then compare
bills. The one who has the bill with the highest
denomination wins all three bills. What is the
probability that Rudolph will win? Thomas?
Neil?
e) What is the expected value for each of the
three players of the game described in d)?
C.60. @ A man plays the following game. He
flips a coin. If it lands on heads, then he wins $1.
If the coin lands on tails, then the man flips again.
If the coin comes up heads on the second flip,
then the man wins $2; otherwise he flips again.
He continues flipping in this manner until he
finally obtains a head. If he obtains his first head
on the n^h toss, then he wins 2"-^ dollars.
a) What would be a fair price for the man to
pay for the privilege of playing the game, if he is
allowed at most six tosses?
b) What if he is allowed to continue tossing
until a head comes up, regardless of how long that
takes?
*C.6i. @ Tickets for the Pennsylvania "Big Fifty"
lottery contain a three digit red number, a two
digit white number, and a one digit blue number.
A drawing is held weekly in which the winning
red, white, and blue numbers are randomly
selected. The following prizes are awarded(9):
$50,000 if your red, white and blue
numbers all match the corresponding winning
numbers;
$1,000 if your red and white numbers
both match the winning numbers;
$100 if your red and blue numbers both
match;
$25 if your white and blue numbers both
match;
$25 if your red number matches;
$5 if your white number matches.
You may only vsdn one prize per ticket.
In addition, if you do not win one of the above
prizes and if your blue number matches the
winning blue number, then your ticket is entered
into a special drawing, which is held after twenty
million tickets have been sold in all. This drawing
awards one prize in each of the following
amounts: $1,000,000, $25,000, $20,000,
$15,000, and $10,000.
If you buy one ticket for $.50, what is your
expected value?
Miscellaneous
C.62. § § In how many ways may the numbers
1, 2, 3, ..., 8 be placed in the eight cells in Figure
C.ii so that no number is directly to the left of or
directly above a smaller number?
FIGURE C.ii
C.63. |h| § In how many ways may the numbers
1, 2, ..., 9 be placed in the nine cells in Figure C.12
so that no number is to the left of or above a
smaller number?
FIGURE C.12
C.64. |h| § The butcher left his shop for only a
few moments. However, while he was gone.
Beggar, the neighborhood mutt, wandered in and
found nine sausages hanging as pictured in Figure
C.13.
In how many different ways could Beggar
possibly eat all nine sausages, one at a time, if he
never eats a sausage higher on a string unless he
has finished all sausages below it.
^^^: There is also an additional chance of four in one
million of winning a Cadillac or Lincoln; but, to simplify
the problem, we ignore this possibility.

374
APPENDIX C
FIGURE C.13
C.65. § How many routes are there from A to B
in the diagram in Figure C.14 if you can only
travel toward the South or the East?
A
FIGURE C.15
C.67. |h| § Andy Hardly lives on the outskirts
(the corner of A Street and First Avenue) of the
town of Planesville. He attends Planesville High
which is located at the intersection of two of the
streets pictured in Figure C.16. Because Andy
likes variety, he plans to follow a different route
each day for as long as he can. However, to avoid
backtracking, he always walks in a southerly,
easterly, or south easterly direction. He has
computed that he will be able to select different
route in this manner for exactly 377 days.
Where is the school located?
Andy Hardly's house
FIGURE C.14
C.66. § @ Our local mailman is only willing to
walk in a southerly, easterly or southeasterly
direction. In the diagram in Figure C.15 (which
represents our town), how many routes are there
for the mailman to walk
a) from A to B?
b) from A to C?
c) from A to D?
d) from A to E?
e) from A to F?
First Avenue
Second Avenue
Third Avenue
Fourth Avenue
Fi<th Avenua
Sixth Avenue
St St St St St
FIGURE C.16

PROBABILITY
375
C.68. |h| § Using the array in Figure C.17 count
the number of ways in which you could start at a
C, move to an adjacent O, etc., to spell the word
COUNT.
C
COC
COUOC
COUNUOC
COUNTNUOC
COUNUOC
COUOC
COC
C
FIGURE C.17
C.69. US When the first man on Earth was told
that a companion for him would be created, he
made the sign, shown in Figure C.18 - both to
introduce himself and to test Eve's intelligence.
a) In how many ways can you spell "MADAM
I'M ADAM" by starting at an exterior M, moving
horizontally or vertically to an adjacent A, etc.,
and ending at an exterior M?
b) What if you are allowed to start at any M
and end at any M?
M
MAM
MADAM
MADADAM
MADAMADAM
MADAMIMADAM
MADAMADAM
MADADAM
MADAM
MAM
M
C.70. |h| § In how many ways can you start at
an S in the array in Figure C.19 and spell SPHINX
by moving from letter to adjacent letter. (Two
letters are adjacent if they are connected by a line
segment.)
S-P-H-I-N-X
P-H-I-N-X-S-P
H-I-N-X-S-P-H-I
I-N-X-S-P-H-I-N-X
N-X-S-P-H-I-N-X-S-P
X-S-P-H-I-N-X-S-P-H-I
P-H-I-N-X-S-P-H-I-N
I-N-X-S-P-H-I-N-X
X-S-P-H-I-N-X-S
P-H-I-N-X-S-P
I-N-X-S-P-H
FIGURE C.19
FIGURE C.18

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376

BIBLIOGRAPHY 377
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1956.
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19. Gardner, M., The Second Scientific American Book of Mathematical
Puzzles and Diversions, Simon and Schuster, 1961.
20. Gardner, M., New Mathematical Diversions from Scientific
American, Simon and Schuster, New York, 1966.
21. Gardner, M., The Numerology of Dr. Matrix, Scribner's, New
York, 1967.
22. Gardner, M., The Unexpected Hanging and Other Mathematical
Diversions, Simon and Schuster, New York, 1969.
23. Gardner, M., Martin Gardner's Sixth Book of Mathematical Games
from Scientific American, W. H. Freeman and Company, San
Francisco, 1971.
24. Gardner, M., Mathematical Carnival, Knopf, New York, 1975.
25. Gardner, M., Mathematical Magic Show, Knopf, New York, 1977.
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27. Gottlieb, A., "Puzzle Corner" column. Technology Review, Vol.
69- (1966- ).
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Dover, New York, 1959.
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to Solve Them, Dover, New York, 1972.
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Dunn, Van Nostrand Reinhold, New York, 1971. Some of the
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Dunn, Dover, New York, 1980.

378 BIBLIOGRAPHY
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1968.
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Communications/Research/Machines, 1973.
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York, 1957.

Hints and Solutions
CHAPTER 1
Exercises
1.1. Hint: Sue's mother must be one of the five
women whose names are given.
1.4. Solution: The first step is to make a chart,
listing the players* initials in the left column and the
positions along the top. We then use the given
information to place v^s and X's in the chart. Using
the clues, we can immediately place X's in some
boxes of the chart. For example, clue (a) enables us
to place an X in the box where Allen's row meets the
catcher's column. To indicate that clue (a) was used
to place this X, we use a subscript (Xa). We continue
eliminating boxes in this manner:
1. Allen is not the catcher (clue a).
Ed is not the second baseman (clue b).
Harry is not the third baseman (clue d).
Neither Paul nor Chuck is the pitcher (clue e).
Ed is not an outfielder (clue f).
Chuck, Harry, and Allen are not outfielders
(clue h).
Sam is not an infielder or in the battery
(clue h).
Neither Paul nor Allen is the shortstop (clue i).
Neither Paul, Harry, nor Bill is the second
baseman or the catcher (clue j).
Neither Ed, Paul, nor Jerry is the right fielder or
the center fielder (clue m).
Bill is not the shortstop or the third baseman
(clue n).
At this point, our chart appears as follows:
Allen
Bill
Chuck
Ed
Harry
Jerry
Mike
Paul
Sam
P
X.
X.
Xh
c
X.
Xi
X,
X,
Xh
lb
Xh
2b
X,
Xb
X,
X,
Xh
ss
X,
x„
X.
Xh
3b
x„
Xd
Xh
rf
Xh
Xh
Xf
Xh
Xni
Xni
cf
Xh
Xh
Xf
Xh
X.
Xm
If
Xh
Xh
Xf
Xh
We cannot tell yet who plays what position, so we
return to the clues to make a list of who is married,
and so on:
380

HINTS AND SOLUTIONS
381
The second baseman is engaged (b).
The pitcher is married (g).
Sam is married (k).
The catcher and the third baseman are not
bachelors (1).
Ed, Paul, Jerry, the right fielder, and the center
fielder are bachelors and everyone else is married (m).
From these observations, we can conclude that:
2. Ed, Paul, and Jerry are not the pitcher, the
catcher, or the third baseman. (Place X2 in the
boxes where rows Ed, Paul, and Jerry meet
columns p, c, and 3b.)
3. Sam is not the right fielder or the center
fielder. (Place X,'s.)
4. The only possible position remaining for Sam
is left field. (Place J,.)
5. No one else can be the left fielder (X,'s).
6. The only possibility remaining for Paul is
first base (v/J.
7. No one else can be the first baseman (X/s).
8. The only possibility for Ed is now shortstop
9. No one else can be the shortstop (X^'s).
10. The only possibility remaining for Harry is
pitcher (y,J.
11. No one else can be the pitcher (X,/s).
12. The only possible position remaining for
Jerry is second base (/,2)-
13. No one else can be the second baseman
(x,;s).
14. The only position remaining for Allen is
third base (/j^).
15. No one else is the third baseman (X,,'s).
16. The only position remaining for Chuck is
catcher (>/,(,).
17. No one else is the catcher (X,^).
18. This leaves Mike and Bill as the right
fielder and the center fielder. From clues (c) and
(o), Mike must be the right fielder, leaving Bill
to be the center fielder.
This completes the chart:
Allen
Bill
Chuck
Ed
Harry
Jerry
Mike
Paul
Sam
P
Xn
Xn
Xe
X2
Ao
X2
Xn
Xe
Xh
c
X3
X,
A.
X2
X,
X2
Xn
X,
Xh
lb
X7
X7
X7
X7
X7
X7
X7
V.
Xh
2b
X.3
X,
Xn
X,
X,
V.
Xn
X,
Xh
ss
X,
Xn
X,
K
X9
X,
X.
X,
Xh
3b
Vm
Xn
Xn
X2
X,
X2
Xn
X2
Xh
rf
Xh
X18
Xh
X,
Xh
Xn,
Vn
Xn,
X3
cf
Xh
Vn
Xh
X,
Xh
X.
X18
Xn,
X3
If
Xh
X3
Xh
Xf
Xh
X5
X5
X5
V.
Note that we have entered all the information
about the problem in this one chart, with subscripts
indicating the sequence in which the various /'s and
X's were entered. Without these subscripts, there
would be no way to reconstruct our reasoning from
the final chart alone. Even with the subscripts,
reconstructing the argument from the chart is a
tedious task; thus the verbal description of the steps
in our solution is important for others to be able
to follow our argument. Bear this in mind if you
ever write up a solution for someone else to look at.
You should include a verbal description of your
procedure as well as charts showing several
intermediate points along the way.
1.5. Hint: Make a chart showing what happens
each day. Whom can St. Jacques play on the fourth
day?
1.6. Hint: Make a chart with columns as the nine
days of the vacation and the rows as the three noise-
makers.
1.7. Hint: Make a chart in which the column
headings are the various events and the row headings
are the names of the teams. Include a column for
total points earned by each team. How many points
were scored altogether? How many could each team
have obtained?
1.8. Hint: Note that a total of ten points is scored in
each game. What were the scores of the three games in
which Alice participated ? (Caution—Alice outscored

382
HINTS AND SOLUTIONS
her opponents by 22 points, but this does not mean
that she scored only 22 points.)
1.9. Comment: Note that each of clues 1-4 tells
us only that the particular resorts mentioned have
particular features. They do not imply that the other
resorts do not have these features. Thus, for
example, we cannot conclude from clue 1 that the
Shangri La does not have a swimming pool. Etc.
1.11. Hint: First consider the characteristic of
knowing the odds. Since everyone is sitting next to
someone who does know the odds, and yet four of
the five are sitting next to someone who does not
know the odds, the only possible seating arrangement
is to have three people who know the odds sitting
next to each other, as this figure indicates:
NO ^O
O = knows the odds
NO = doesn't know the odds
Do the same for each characteristic. Then consider
Edie first and Cleo next.
1.13. Hint: Make two charts—one with three
columns representing the caller, the person he
intended to call, and the person who was actually
reached; and the other plotting the names of the
men against their occupations.
1.14. Solution: He said to himself: "If my face
were clean, then Jack would realize that his own
face is dirty, for why else would Jill be laughing.
Since Jack did not draw this conclusion, my face
must be dirty too."
1.15. Hint: See the solution of Exercise 1.14.
1.16. Hint: Don't forget that Carla knows that
John did not receive what he ordered. Also, see the
solution of 1.17 below.
1.17. Solution: Each cup contains five coins. There
are only six possibilities for the contents of a cup:
(Let N represent one nickel and D represent one
dime.)
NNNNN
(25c)
NNDDD
(40 c)
NNNND
(30c)
NDDDD
(45c)
NNNDD
(35c)
DDDDD
(50c)
Since there are six cups, each of the above
possibilities must actually occur—one in each cup.
We begin by making a chart showing what each
of the first two men could have felt (based on their
statements), the possible incorrect labels that could
have led them to these conclusions, and what their
cups could actually have contained (see the figure).
felt
NNND
Oil man
NDDD
second ^^^^
oilman dddo
label on cup
30C
35C
40C
30c
50c
actual contents
NNNDD =35C
NNDDD = 40C
NDDDD = 45c
NNNNN = 25C
NDDDD = 45C
For example, if the first oil man felt NNND, the
only way he could know that the fifth coin was a
dime is if the label said 30c.
Now consider the third man. On the basis of
feeling only two coins, no matter what they were,
there are still four possibilities for what his cup
could contain. He could eliminate one of these by
knowing the label on his cup, but the only way that
he can eliminate more than one is if he can figure
out what the first two have in their cups.
If the label on the third man's cup said 50c, he
could know all about the second man's cup (it would
be labeled 30c and would contain NNNNN), but he
would not be able to tell anything about the first's.
Similarly, if the third man's label said 45c, 40c,
35 c, or 25 c, he would not have been able to make
his claim.
However, if his label actually said 30c, then the
second cup must have been labeled 50c and must
have contained NDDDD. But then, the first cup
could not have contained NDDDD and could not
have been labeled 30c. It must therefore have been
labeled 35c and must have contained NNDDD =
40c, as this figure shows:
oil man
first
second
third
label
35C
50c
30c
actual contents
NNDDD = 40C
NDDDD = 45C

HINTS AND SOLUTIONS
383
If the third man felt DN, he could now conclude
that he has NNNDD. This is the only case that
works.
We are now ready to consider the fourth leader of
the oil industry. As he felt only one coin, there seem
to be five possibilities for the contents of his cup.
However, he is capable of the same reasoning we
used above, and so he can deduce the labels and
contents of the cups of the first three fellow students.
Using the fact that the label on his cup is incorrect,
he may be able to eliminate four of the five
possibilities.
If he felt a dime, then no matter what his own
label read, he would only have been able to eliminate
three of the possibilities. (Check this.) But to know
what he has, he must be able to eliminate four.
Hence, he must have felt a nickel. The only way he
would have been able to come to a conclusion is if
his cup were labeled 25 c, in which case it would
contain NNNND. We obtain the chart below:
oil man
first
second
third
fourth
label
35C
30c
25C
actual contents
NNDDD = 40C
NDDDD = 45c
NNNDD = 35C
NNNND = 30C
Thus, the two remaining cups must be labeled 40c
and 45 c; one must contain five nickels and the other
five dimes. They thus contain 75c between them.
1.20. Hint: Set up a chart with five columns
representing the competitor's name, school, entry number,
prior standing, and position in the high jump. Enter
as much information as possible, using letters, x, jy,
0, etc., to represent unknown quantities. For
example, since Manners' position in the high jump
was one number higher than his prior standing,
make the following entry in one row of the chart:
competitor
Manners
school
entry
number
prior
standing
X
high
jump
X + 1
Be sure you use different letters to represent
different unknown quantities.
1.21. Hint: Which of the five houses is the red
house? Consider cases.
1.22. Hint: First find out which show is at which
theater. Make a chart showing all possible pairings
as the row headings, and the shows and days of the
week as the column headings. (There are twenty-five
rows and ten columns.) Fill in what you can, then go
back to the beginning and go through the clues
again.
1.23. Hint: Hook A and B together, C and D
together, and proceed as in Problem 1.6 (page 22).
Then hook B, L, and C together and proceed as in
Problem 1.6.
CHAPTER 2
Exercises
2.3. Hint: What is the largest number who could
possibly be Truthfuls?
2.4. Solution: How many Truthfuls could there
be? None? No, because then the fourth troll would
be telling the truth and hence would be a Truthful.
One? Possibly, if the fourth troll is the only Truthful.
Two? Possibly, if the third and fifth trolls are
Truthfuls. Three? Possibly, if the first, second, and
fifth trolls are Truthfuls. Four? No, since the first
three would be lying. Five? No, since the first three
would be lying.
Consider, now, what the fifth troll might have
said. Remember that his answer left no doubt in
Silas' mind as to the clan affiliation of each.
If he said "three," then Silas would still have
been in doubt because the fourth troll could have
been the only Truthful or the first, second, and
fifth trolls could be telling the truth.
Similarly, if the fifth troll said "two," two possible
cases could remain (only the fourth is a Truthful,
or the third and fifth are both Truthfuls).
If the fifth says anything else, then it is no longer
possible that there are either two or three Truthfuls
(see the cases above). Thus, the fourth must be the
only Truthful.
2.6. Hint: What kind of troll could say about
himself that he is a Liar?

384
HINTS AND SOLUTIONS
2.8. Hint: Consider the possible alternatives for
Hocus (he is a Truthful; he is a Liar; he is an
Alternator who first tells the truth; he is an Alternator
who first tells a lie). In each case, see what Pocus
can be.
2.11. Hint: Ask a hypothetical question within a
question. For example, " If I were to ask ..., what
would you say?"
2.15. Solution: We will refer to the statements as
Jeeves 1, Jeeves 2, etc. Jeeves made 3 statements
(1, 2, 4) implying his innocence. At least one of these
must be true. Therefore, Jeeves didn't do it.
Similarly, Jessica 1, 3, 4 imply she was innocent.
We now know that certain statements are true and
certain are false. These are shown in the chart below:
Thus Jessica was not in Chicago. Our chart is now:
Jeeves
12 3 4
T F
Fifi
12 3 4
Julia
12 3 4
F
Jessica
12 3 4
T
Since everyone refers to Jessica's being
blackmailed, let's consider the possibilities: She was
being blackmailed, or she wasn't being blackmailed.
Case 1 Suppose she wasn't being blackmailed.
Then her first two statements are true and
therefore the last two are false. Therefore Fifi is innocent
(Jessica 4 is F) and Julia must be the murderess.
But in this case, Julia 1, 2, 3 are F. This contradicts
the fact that each person makes two true and two
false statements.
Case 2 Jessica was being blackmailed. This enables
us to further fill in the chart:
Jeeves
12 3 4
T F
Fifi
12 3 4
T
Julia
12 3 4
T F
Jessica
12 3 4
T F
Now make a secondary assumption considering
whether or not Jessica was in Chicago. If she was,
then Jessica 1 and 3 are true, forcing Jessica 4 to be
false, so that Fifi is innocent and Julia is the
murderess. But then, Julia 1, 3, and 4 are false, again
giving a contradiction.
Jeeves
12 3 4
T F
Fifi
12 3 4
T
Julia
12 3 4
T F
Jessica
12 3 4
TF FT
Note Jessica 4 must be true, and Fifi is the
murderess. (We still have to check that this answer is
consistent with the given information.)
We can fill in the chart as follows:
Jeeves
12 3 4
T F F
Fifi
12 3 4
F F TT
Julia
12 3 4
TT F F
Jessica
12 3 4
TF F T
Jeeves 3 must therefore be true and then
everything fits. Hence, Fifi is the murderess.
2.20. Solution: Label the statements as Earl 1,
Earl 2, etc. From Earl 2, Earl is not the architect
since, if he were, this statement could not possibly
be true.
Similarly, Luis is not the architect by Luis 1,
and Randy is not the mason by Randy 2.
Since we now know that neither Earl nor Luis
is the architect, we have two possibilities: Case 1,
Moe is the architect; Case 2, Randy is the architect.
Case 1 In this case, Moe 2 must be true so that
Luis is the mason. This leaves Earl and Randy to
be the plumber and the carpenter (not necessarily
in that order).
However, in this case. Earl 1 and Randy 1 would
both have to be true; but they contradict each other.
Thus, this case leads to a contradiction.
Therefore Case 2 holds: Randy is the architect.
(Randy's statements will now be ignored as neither
need be true. He does not mention the architect.)
Moe cannot be the mason since if Moe 2 were true,
Luis would have to be the mason—contradiction.
Therefore Moe is the carpenter or the plumber
and Moe 1 must be true. Therefore Earl 1 is false
and Earl cannot be the carpenter or the plumber,
so Earl is the mason.
Similarly, Luis 2 is false, so Luis is not the
carpenter. This leaves Luis to be the plumber and
Moe to be the carpenter.

HINTS AND SOLUTIONS
385
2.23. Hint: Can there be more red cards in the top
half than there are black cards in the bottom half?
2.24. Solution: We begin by representing the
statements symbolically. Let g, h, and / represent the
following statements:
g: Martians are green.
h: Martians have three heads.
/: Martians can fly.
Then the astronaut's statement, which we are told is
true, can be represented as
By Observation 9 (page 48), either k -»(/z v (~/))]
is false or [{g <^f) /\{^h)] is true.
Consider cases:
Case 1 \g^{hw{^f))] is false. Then, by
Observation 10, g must be true and {h v ( ~/)) must
be false. That is, g is true and h is false and / is
true (by Observation 5).
Case 2 [{g <-►/) a ( ~/z)] is true. Then, by
Observations 3 and 11, /z is false and g and / have the same
truth values. However, since Martians have at least
one of the three characteristics, it is not possible for
^, /, and h all to be false. Thus, g and / must both
be true, with h false.
Since both cases lead to the same conclusion, we
find that Martians do not have three heads, they are
green, and they can fly.
2.28. Hint: Consider whether Archie owns the
Chrysler or the Chevy. Caution—if Archie does own
the Chevy, the second clue tells us nothing about the
color of the car.
2.30. Hint: Express this problem symbolically.
2.31. Hint: Express this problem symbolically.
2.32. Hint: Express this problem symbolically.
CHAPTER 3
Exercises
3.2. Solution: The easiest way to solve the problem
is to work backward. However, we will take the
algebraic approach:
Let M = the amount of dollars they started with.
Dinner cost \M.
This left them with M - \M = \M.
The two theater tickets cost $19.60.
This left them with \M - 19.60.
The taxi cost J {\M - 19.60).
This left them with I {\M - 19.60) = ^Af - 14.70.
The nightclub cost $23.10.
This left them with JM - 14.70 - 23.10 =
iM - 37.80.
The cab fare home cost '2(2^ - 37.80) =
\M - 18.90.
This left them \M - 18.90.
After the $1 tip they are left with JAf - 18.90 - 1,
which must = 4.10.
Therefore we get the equation
\M - 19.90 = 4.10
\M = 24.00
M = 96.
Since they started with $96 and ended with $4.10,
they spent $91.90.
3.4. Hint: Let C = the amount of candy.
How much did Samuel take and what was left?
How much did Mildred take and what was left ?
3.5. Solution: Each share is 4 bottles, so Tom
gives Mack 3 bottles and Don gives Mack 1. Hence
they should split Mack's money in the ratio of 3 : 1.
3x + X = 8.40
x = 2.10
Tom gets 3x = $6.30 and Don gets x = $2.10.
3.7. Hint: s = the number of sacks of wheat seed,
s is also the price per sack, so that altogether Farmer
Grey spent s^ on wheat seed; etc.
3.10. Solution: Assuming that x is well defined,
then, squaring both sides,
x^ = 1 + Vl + yi + ^/^
But the right side is just 1 + x
x^ = \ + X
x' - X - \ =0
1
by the quadratic formula.

386
HINTS AND SOLUTIONS
But
y^
solution is x
is negative and x is positive, so the
_ 1 + v^
3.11. Hint: See the solution of Exercise 3.10.
3.14. Hint: Let n = the number of $.03 screws
that Calvin bought. Observe that n is also equal to
the number of $.04 screws, etc. Let>; = total amount
that each person spent.
How much did Wendy spend on $.03 screws?
How many $.03 screws did she buy? Do the same
for the $.04 and $.05 screws.
3.15. Solution: Define A, B, C, D, E, and F in the
obvious manner:
(\) A + B = 647
(2) B + C = 675
(3) C + D = 599
(4) D + E= 583
(5) F = 370
(6)A + B+C + D-hE + F= 1927
From (5), A + B + C + D + E= 1557.
From (l)and (3),
(A + B) + (C + D) = 647 + 599 = 1246
E= 1557 - 1246 = 311.
Now substitute back in (4) to find D, and
substitute back in (3), to find C, etc., getting A = 299,
B = 348, C = 327, D = 272, £=311.
3.25. Solution: Let S = Sue's present age
C = Chin's present age
Make a chart as shown, with row headings for Sue
and Chin, and column headings for present, past,
and future. Since Sue is 3 times as old as Chin was
when Sue was as old as Chin is now, we can put
Present
Sue 5
Chin C
Past
Future
C in the Sue-past box and - in the Chin-past box.
Also, since Sue will be 56 when Chin is as old as Sue
is now, we can put 56 in the Sue-future box and S
in the Chin-future box. The result is:
Present
Sue 5
Chin C
Past
C
S
3
Future
56
5
Since the difference between Sue's and Chin's ages
remains constant,
S
C -
56 - S.
(1)5-
We obtain the following system of equations:
3
(2) 5 - C = 56 - 5.
(T) 35- 3C= 3C- 5
giving
(1") 25- 3C = 0
(2) 25 - C = 56
Subtracting, - 3C + C = - 56
2C = 56
C = 28, 5 = 42.
3.26. Hint: See the solution of Exercise 3.25.
3.28. Hint: See the solution of Exercise 3.25.
3.32. Solution: Imagine a flag at the front of each
train. At the time the trains meet, how far are the
flags apart? Answer, 0 ft. At the time the trains just
pass each other, how far are the flags apart? Answer
270 + 192 = 462 ft, as the figure shows.
Therefore the total distance traveled is 462 ft. At
what rate are the flags separating? In each hour, one
flag moves 45 mi and the other moves 60 mi. So
they separate at the rate of 105 mph. To maintain
consistency of units, we convert to feet per second:
88
105 mi/hr = — • 105 ft/sec.
60
d
Using the formula d = rt or t = - we get
r
462 ft
"88 • 105 ft
3 sec.
60
sec

HINTS AND SOLUTIONS
387
A slightly different approach is as follows (note
that the time interval is the same for both trains):
For the first train,
, 45 • 88
17 u ^ • ^60-88
For the second tram, d^ = t
60
where d^ + d2 = the distance between the flags
= 462 ft.
So we add the two equations to obtain
^^o _, _, 45-88 60-88
462 = d, + d2 = t + t
60 60
105 • 88
462 = t
60
t = 3 sec.
3.33. Hint: If ci = the distance between Topeka
and Santa Fe, what are the rates of the two trains ?
3.35. Hint: What is the relationship between the
distance from Miami to the point at which the trains
meet and the distance from Washington to that same
point? What is the ratio of the rates of the two
trains ?
3.37. Hint: When will the trains collide?
3.38. Solution: Picture the action as shown in the
figure.
1
Nf-
150 yd
-+-
100 yd
HS
(a) Solution by two methods:
Let r, = the rate of the first cyclist
r2 = the rate of the second cyclist
d = the length of track
fi = the time from the start until they meet
the first time
f2 = the time from start until they meet the
second time.
Method I
r, r^t^ 150
^2 r^ t,
r^ _ r, f2 _
^2 r^t^
150
d- 150
300(i - 15,000 =
^ - 350(i =
d{d - 350) =
d=^
(Note: 0 is extraneous.
Method II
(r, + rjf,
('•i + '•2)?2
(^ + r,)r,
('•i + '•2)^2
3r,
d-h 100
d
(b) r, 150
d- 150
d-h 100
2d - 100
d-h 100
2d - 100
d" - 50d ~ 15,000
0
0
0, 350.
.)
= d
= 3d
d 1
~3d~3
= h
= r,t2
= 3rj^
= 3-150
= 450
= 350.
150 3
r^ d - 150 200 4
3
Therefore r2 > r^ and the second cyclist must win.
(c) Observe that in time f, the first cyclist goes
150 yd and the second goes 200 yd; so in each
time interval f^, the second cyclist gains 50 yd. To
overtake his opponent, the faster cyclist must gain
a total of 350 yd. This will take 7 time intervals
ti. At the end of the seventh time interval fi, the
first cyclist will have gone 7 • 150 = 1050 yd = 3
laps of the track; thus, the winner will overtake
the loser at the southern tip of the track.

388
HINTS AND SOLUTIONS
3.39. Hint: Let n = the number of chairs showing
on each side of the lift at one time.
Express the time of the skier's trip in two different
ways: once using the up chairs; once using the down.
3.4L Hint: The movement of the boat relative to
the driftwood is independent of the rate of the
current. Thus, you can find the rate of the current
even though you do not have enough information to
determine the rate of the boat.
3.42. Solution: Let s = the number of steps
showing at one time, t = the time it took Irv to take
3 steps (and Liz 2), and r = the number of steps
that disappeared during time f.
Since Liz reached the top in 24 steps, it took her
12f seconds. During this time, 12r steps disappeared,
so that Liz actually covered 5 - 12r steps. That is,
24 = 5 - 12r.
Similarly, consider the number of steps Irv
climbed: 30 = 5 - lOr.
Solving simultaneously,
6 = 2r
3 = r
s = 60.
3.43. Hint: See the solution of Exercise 3.42.
3.44. Solution: Let t be the total time in hours that
Flog rides and let T be the total time that he walks.
Then he covers a total distance of 56f + 47 in a total
time of f + r. Since each of his companions must
cover the same distance in the same time, each rides
for time t and walks for T. During the first
passenger's ride, the distance covered by the first
passenger is 56f km, while the walkers cover a
distance of 4f km (see Figure 1).
During the motorcycle's return trip, the cyclist
and the walkers are approaching each other at a rate
of 56 + 4 = 60 km/hr. As they are 56f - 4f = 52f km
. 52f
apart, the return trip of the motorcycle takes — hr.
During this time, all walkers (including the first
passenger) cover a distance of 4j—| km (Figure 2).
The distance between the two groups of walkers
remains 52f km.
The second passenger now rides for time r, exactly
overtaking the first passenger. The motorcycle
returns I time — 1. The third passenger rides for
/ . 52f\ , ^
time t; the motorcycle returns time —I for the
->
FIGURE 1
FIGURE 2
START h-
START V
distance in
kilometers
41
52t
FIGURE 3

HINTS AND SOLUTIONS
389
final walker. At this point the situation is as indicated
in Figure 3.
The last passenger rides for t hours and catches up
at (78^)f km; the total time for this trip is
52f 33
4r + 3 • — = — t.
60 5
Since the trip must be completed in H hr = | hr,
33 5
— r < -
5 ~ 3
25
f < — .
-99
On the other hand, since the total distance must be
at least 19.3 km,
78^, t > 19.3
19.3 193
t > ■
(78^) 784 ■
That is, f must satisfy
There are values of t that satisfy both inequalities,
for example.
Therefore, the trip is possible. For instance, if
f = i hr = 15 min, each person walks for
3r + 3
(-1 =
\ 60 /
45 + 39 = 84 min
and rides for 15 min, giving a total trip of 99 min.
In this time they cover
84^) + 15(,^) = \9i = 19.6 km
(more than is needed).
3.45. Hint: Let n = the number of stations at which
the EE stops from the time it leaves Continental
Avenue until it reaches 34th St. (including 34th St.).
Let w = the number of stations after Continental
Avenue up to and including Queen's Plaza. Let
7?F = the rate of the F train (in stations per minute)
and 7?EE = the rate of the EE train (in stations
per minute).
3.52. Hint: Let A = the initial amount of grass in
the field (volume),
C = the number of days required
for a cow to eat A,
H = the number of days required
for a horse to eat A,
S = the number of days required
for a sheep to eat A,
G = the number of days required
for uneaten grass to grow A.
3.54. Hint: If 5 = the number of shirts that Leroy
bought, then how much did each shirt cost and what
was the total amount of money that he spent on
shirts?
3.55. Solution: Although this can be done
algebraically, it is quickly answered if you observe that the
amount of vinegar that was added to the oil must
exactly equal the net amount of oil that had been
removed. Since the oil that was no longer in the oil
bowl was now in the other bowl, the amount of
vinegar in the oil and the amount of oil in the
vinegar were equal.
3.60. Hint: What is the total amount that the five
spent? How much did Bruce and Cookie spend
together? How much did Chuck spend?
3.61. Hint: Let L = the length of the left arm and
R = the length of the right arm. Let T = the weight
of 6 tomatoes, B = the weight of the bag of beans.
Note that you will be able to determine — but not
R and L separately.
3.71. Hint: (x - y){x -\-y) = x^ - y.
3.72. Hint: Let n = the selected number.
After the deck is shuffled, there are n cards face up
and 52 - « cards face down.
In the pile of n cards taken from the top of the
shuffled deck, let / = the number of cards that are
face up. How many cards in the packet are face down ?
How many cards in the rest of the deck are face up ?
3.73. Solution: In a pile of cards constructed as
in this problem, if the first card has value V, how
many cards are in the pile? To reach 12 we must add
12 - K more cards. So altogether there are 12 — K
(added cards) + 1 (the original card) = 12 - V +
1 = 13 - K. We can also look at this in another
way: The pile has cards corresponding to numbers
V up to 12; the numbers from 1 to V - \ are
omitted. Therefore, there are 12-(K-1) =

390
HINTS AND SOLUTIONS
13 - V cards in the pile. These two approaches are
illustrated below:
K, + K, +
1st card
12-V cards <
value V
V + 1
12 = V + (12 - V)
total is
I + (\2 - V) = 13-V cards
V-l cards
12-(V-1)
13-V cards
V
V+l
12
> 12 cards
+ K^= 13(6-4)+ 5
= 26 + 5 = 31.
3.74. Hint: Could the selected card have been
placed anywhere in the deck, or does the magician
make sure that it is in a certain position? What
position ?
See also the solution to Exercise 3.73.
3.76. Solution: Let x = the number of cards taken
by the first person, y = the number of cards taken
by the second person. Note that y > x and
X + jy < 26. The magician is left with 52 - x - y
cards.
In the face up pile of 26 cards dealt by the
magician on the table, the card noted by the first
subject is the xth from the bottom and the card
noted by the second subject is the yth from the
bottom. See the illustration.
Cards face up
26
y (2nd person's card)
X (1st person's card)
The trick can now be explained as follows:
Suppose there are n piles and the top cards have
values K„ K^,..., V„.
Then the first pile has 13 - K^ cards,
the second pile has 13-^2 cards,
the «th pile has \3 - V„ cards.
Therefore there are (13 - KJ -I- • + (13 - V„) =
13« - (Ki + K2 + • • + K„) cards on the table.
There are R cards left over and altogether this will
equal 52. That is,
52 = 13« - (K, + K2 + • • • + V„) + R
or equivalently,
K, + K2 + • • • + K„ = 13« + 7? - 52.
Since 52 = (13)(4),
K, + K2 + • • • + K„ = 13(« - 4) + 7?.
In the particular problem, « = 6 and the number of
cards left over is 5
When the 26 cards are turned face up on the table,
the magician has 52-x-y-26 = 26-x-y
cards left in his hand. When the cards are placed
face down at the bottom of his packet, the first
subject's card is at the 26-x-y + x=26- yth
position from the top. Similarly, the second subject's
card is at the 26 - xth position (see the illustration
at the top of page 351). Altogether, the magician is
holding 52 - X - y cards, hence the 26 - xth card
from the top is the 52 - x - >; - (26 - x) + 1 st
card = the 27 - jyth card from the bottom.
Throwing away the bottom card (which is the
bottom card of the bottom half of the deck) leaves
the selected cards at the 26 - ^'th position from the
top and the 26 - ^'th position from the bottom.
Hence, when the bottom half is turned over, the two
selected cards will both appear at the 26 - jyth
position.
Note that since x <y, -x> -y, and 26-
X >26 - y.

HINTS AND SOLUTIONS
391
Cards face down
52-X-3; <
26-x-y
26 <
- 1
- X ^
-y
-y^\
26
I 26-x-3; + x ■■
26-y
Therefore 26 - jy < half of the deck in the
magician's hand.
3.77. Hint: Let x = the number of cards that are
removed, y = the chosen number. Use a counting
argument similar to the one given above for the
solution of Exercise 3.76 to determine the position
of the selected card.
3.78. Hint: Determine the position of the selected
card by an argument similar to the one used above
in the solution of Exercise 3.76.
3.79. Hint: Did the magician really select the four
of diamonds? Where was the four of diamonds
when the trick started?
Work backward and find out, then justify your
answer.
> 26-x
> 52-x-y-(26-x) = 26-3;
by 2 is the same as the remainder when the last
digit is divided by 2—it is 1 if the last digit is odd
and 0 otherwise.
Similarly,
ABCDEF = F (mod 5).
Thus, the remainder when ABCDEF is divided by 5
is the same as the remainder when F is divided by 5.
For example,
2387 =7=2 (mod 5).
To consider remainders upon division by 3, we
use the fact that any number is congruent modulo 9
to the sum of its digits, and, by Theorem 4.9, this
is also true modulo 3. In particular,
ABCDEF =A + B + C + D + E + F (mod 9),
CHAPTER 4
Practice Problems
4.M 5. Solution: Since
ABCDEF = A • 10^ + B • 10^ + C • 10^ + D • 10^
+ E • 10 + F • 1,
ABCDEF = F (mod 10).
Therefore, by Theorem 4.9,
ABCDEF = F (mod 2).
In general, the remainder when a number is divided
ABCDEF =A + B + C + D + E + F (mod 3).
This enables us to "cast out threes.'* Thus, the
remainder when a number is divided by 3 is the same
as the remainder when the sum of the digits is
divided by 3.
For example,
283147 =2 + 8 + 3+1+4 + 7 = 25 =2 + 5
= 7 = 1 (mod 3),
so the remainder when 283147 is divided by 3 is 1.
To find the remainder when ABCDEF is divided
by 4, observe that 10^ = 100 = 0 (mod 4) and hence
that 10* = 0 (mod 4) for k > 2. Therefore
ABCDEF = lOE + F = EF (mod 4).

392
HINTS AND SOLUTIONS
That is, any number is congruent modulo 4 to its
last two digits. Hence, to find the remainder when
a number is divided by 4, it is only necessary to
find the remainder when the number formed by the
last two digits is divided by 4.
For example, 1873954618 = 18 = 2 (mod 4); so
the remainder is 2 when 1873954618 is divided by 4.
The rule for 8 is obtained similarly:
10^ = 1000 = 0 (mod 8),
so
ABCDEF = DEF (mod 8).
That is, the remainder when a number is divided by
8 is the same as the remainder when the number
formed by the last three digits is divided by 8.
The rule for the number 11 is obtained in a manner
similar to the rule for 9:
Therefore
ABCDEF =
=
=
10
10^
10^
etc.
A • 10'
+ E- ;
-A +
= -1 (mod 11)
= 1 (mod 11)
= - 1 (mod 11)
' + B • 10^ + C • 10^ + D •
10 + F
10^
B-C + D-E + F(mod 11)
(F + D + B) - (E + C + A).
That is, to find the remainder when a number is
divided by 11, first find the sum of the 1st, 3rd, 5th,
etc. digits from the right and then subtract the sum
of the 2nd, 4th, 6th, etc. digits from the right. The
number obtained in this manner will have the same
remainder upon division by 11 as the original
number.
For example,
1879543765 =(5 + 7 + 4 + 9 + 8)
-(6 + 3 + 5 + 7+1) (mod 11)
= 33 - 22 = 11 =0 (mod 11).
So the remainder is 0 when 1879543765 is divided
by 11; that is, 11 divides 1879543765.
Similarly,
29083 = (3 + 0 + 2) - (8 + 9) :
= 10 (mod 11).
17 =
12
That is, the remainder is 10 when 29083 is divided
by 11.
The rule for the number 7 is somewhat more
complicated. It is based on the fact that 1001 =
7 • 11 • 13 =0 (mod 7). Thus
1000 = - 1 (mod 7)
1000000 = (1000)^ = 1 (mod 7)
and so on.
In particular,
ABCDEF = A • 10^ + B • lO-* + C • 10^ + D • 10^
+ E • 10 + F
= (A • 10^ + B • 10 + C) • 10^ + D • 10^
+ E • 10 + F
= (ABC) • 1000 + DEF
= (ABC)( - 1) + DEF (mod 7)
= DEF - ABC.
The general rule for finding the remainder when a
number is divided by 7 is to subdivide the number
in blocks of 3 digits (counting from the right). Add
the rightmost block, subtract the next block, add the
next block, and so on. Divide the resulting number
by 7 to obtain the desired remainder.
For example, to find the remainder when
8137496023 is divided by 7,
8137496023 =023 - 496 + 137-8 (mod 7)
= (23 + 137) - (496 + 8)
= 160 - 504
= - 344 = - 1 = 6 (mod 7).
Therefore, the remainder is 6 when 8137496023 is
divided by 7.
Exercises
4.3. Hint: Is49,999a prime? What would you have
to do to show this?
4.6. Hint: If a number ends in a 0, what can be
said about the prime factorization of this number?
What if it ends in two O's?
What if it ends in three O's? etc.
In the prime factored form of 50!, to what power
is 2 raised? To what power is 5 raised?

HINTS AND SOLUTIONS
393
4.7. Solution: Let B = the number of boys, G =
the number of girls, and Y = the number of years
the parents have been married. Each is a prime
number and they are all distinct.
{B + G)B = 24 + y
Suppose that B is even, that is, B = 2.
Then the left side is even and so Y is even, and
must be 2 (the only even prime is 2). This is
impossible since B and Y must be distinct. Thus
B i= 2 and must be odd.
If G were also odd, then again the left side would
be even and Y = 2, leaving
B(B + G) = 26 = 2 • 13.
In this case, B would have to be 2, which cannot be.
Therefore, G is even (G = 2). The equation
becomes
B(B + 2) = y + 24
B^ + 2B - 24= Y
(B - 4)(B + 6) = y.
But y is a prime and each factor is an integer
since B is. Therefore, one factor must = 1. Since
B - 4 <B + 6
B - 4= \
B = 5
Y= 11.
4.8. Hint: What is the relationship between the
number of times that the /jth switch is reversed and
the number of divisors of /?? How many divisors
must k have for the ^th light to end up on ? Which
lights end up on? (You might want to test your
theory on 10 lights using cards to represent the
lights.)
4.10. Solution: Let ^ = Pp • •/>** be the prime
factored form of a.
Since a^ = ^ is a perfect square, each exponent must
be even, that is, 3r, is even, ?>r^ is even, etc.
Ti is even, r2 is even, etc.
a is a perfect square, that is, a = ^.
Similarly c is a perfect square and c = f^.
25 = c - a=P - e^
= (f-e)(J+e)
To solve this either ^ or
[/■+^ = 25l |/+g=5|
givmg
2/ = 26 or 2/ = 10
/=13 /=5
^ = 12 e = 0
= 192 ^
12^ = 144 and c= 13^
extraneous
169.
Finally, b"- = a^ = 12^ therefore, b = \2^ = 1728.
Similarly, d= W = 2197.
4.11. Hint: Let x, y, and z be the three numbers.
Then
X + y + z = 25
xyz = 540 = 2^ • 33 • 5.
Therefore, x, jy, and z are divisors of 540 which are
less than 25. Note that exactly one of them, say
X, is divisible by 5.
That is, X = 5 or 10 or 15 or 20.
In each of these cases, what can be said about y
and z}
4.12. Hint: Considering that the census taker
knows the number on the door, why did he have to
ask whether the youngest is a twin?
4.13. Hint: If two numbers A and B have the same
remainder when divided by d, what happens when
their difference is divided by d}
4.17. Hint: Write two equations in three unknowns
and eliminate one of the unknowns to obtain one
linear Diophantine equation in two unknowns.
4.19. Solution: Let N = the price (in cents) of one
notebook, L = the price (in cents) of one pencil, and
P = the price (in cents) of one pen (P > 10). Then
the equation representing the amount Garrett spent
is:
695 = 6N + 4L + 5P (1)
and the equation representing the amount Beth
spent is:
285 = 2N + 5L + 3P. (2)
Eliminate N by the addition subtraction method,

394
HINTS AND SOLUTIONS
that is, three times equation 2 minus equation 1 gives
160 = IIL + 4P. (3)
This is a Diophantine equation in two unknowns
and can be solved by the method discussed in the
chapter. Convert equation 3 to a congruence modulo
4.
0 = 3L (mod 4)
Since gcd(3, 4) = 1, we may divide by 3 getting
L =0 (mod 4)
or, equivalently,
L = 4K
(where AT is a positive integer since L must be
positive). Substituting 4K for L in equation 3 we get
160 = 11(4/C) + 4P
P = 40 - \\K.
Since P must be greater than 10, K is restricted to
1 or 2. Therefore, we have two possible cases:
When K = \, L = 4, P = 29;
when K = 2, L = 8, P = \8.
For each case, we substitute the corresponding values
for L and P in equation 1 and find N:
Case 1 6N = 695 - 16 - 145 = 534
N= 89
Case 2 6N = 695 - 32 - 90 = 573
N = 95|.
Since N must be a whole number, the case 2 answer
does not satisfy the conditions of the problem. So
L = 4, P = 29, N = 89, and Wilma paid
1(89) + 1(4) + 1(29) = 122 cents
= $1.22.
4.20. Hint: Set up the equation and solve it by
considering cases—by considering how many
grandfather clocks could have been sold.
4.23. Hint: 3^ = 2 (mod 5).
4.24. Hint: Let x = the first number the subject
selected, and let y = the number the subject
announced. (In the actual example, y = 4.)
If the subject counts "x" as the magician taps the
number y on the clock (the first tap), how many
taps will the magician have made at the time that the
subject counts N (for a general number N)} At this
time, to what number on the clock will the magician
be pointing? (Note that the answer to this question
must be expressed in terms of x, jy, and TV.) What
should N be so that this indicated number will be
"x'* on the clock?
4.25. Hint: Use algebra and digital roots.
4.27. Hint: Use algebra and digital roots.
4.28. Hint: Use digital roots.
4.29. Hint: What is the largest value S could
possibly have?
What is the digital root of 5?
What is the largest value T could have?
What is the digital root of T?
What is the sum of the digits of T?
4.30. Hint: Let G = the number of pieces of gum
in the machine. From the information about Kevin,
G = 1 (mod 4). Therefore, G = 4Ar + 1 for some
positive integer K. Kevin took K + 1 balls, leaving
?>K balls. From the information about Esteban,
?>K = 1 (mod 4). Therefore, K =3 (mod 4), and so
/C = 4/ + 3. Esteban took 3/4-3 gum balls, leaving
97 + 6 gum balls, etc. Remember we are looking for
the smallest possible answer.
4.31. Solution: Let p = the price per plum; then
3/) = the price per apple.
Let A = the total number of apples purchased,
and P = the total number of plums. Then
Pp = 1188, A(3p) = 1188.
Therefore P = 3A, and so A + P = AA. That is,
A + P =0 (mod 4).
The sum of all the baskets = 289 = 1 (mod 4).
Therefore the number of fruits in the remaining
basket is congruent to 1 modulo 4. Of all the baskets,
only two (the one containing 17 fruits and the one
containing 25 fruits) have a number of fruits
congruent to 1 modulo 4.
Consider the two cases:
Case 1 The basket with 25 fruits remains:
/! + P = 289 - 25 = 264

HINTS AND SOLUTIONS
395
4A = 264
A = 66
Theonlysiimofbasketsthatgives66is30 + 17 + 19.
Case 2 The basket with 17 fruits remains:
A + P = 289 - 17 = 272
^A = 272
A = 68
No sum of baskets can equal 68, so this case cannot
occur.
Therefore the basket remaining contains 25 apples,
66 apples cost $11.88 (so each apple costs $. 18), and
the remaining basket is worth $4.50.
4.32. Hint: See the solution to Problem 5 in
Practice Problems 4.M (page 351).
4.33. Hint: See again the solution to Problem 5 in
Practice Problems 4.M (page 351). Observe that
saying that n\m is the same as saying that
w = 0 (mod «).
Note also that a number is divisible by 6 if and
only if it is divisible by 2 and by 3; by 12, if and
only if it is divisible by 4 and by 3; by 14, if and
only if it is divisible by 2 and by 7; by 15, if and
only if it is divisible by 3 and by 5.
The rule for 13 is similar to the rule for 7, since
1001 = 0 (mod 13). The rule for sixteen is analogous
to the rules for two, four, and eight.
4.34. Hint: 99 = 9 • 11.
4.35. Hint: Find the prime factorization of 792.
4.36. (a) Solution: The number of days in 400 years
is:
400(365) + 97 days, where 97 is the number of leap
years in 400 years.
400(365) + 97 = 1 • 1 + 6 = 0 (mod 7).
Thus, 400 years contains an exact number of
weeks, so the day following the 400 year period is the
same day of the week as the first day of the period.
(c) Solution: Label the days of the week as
Sunday—0, Monday—1, Tuesday—2,
Wednesday—3, Thursday—4, Friday—5, Saturday—6. In
order to do this problem, we need a reference day
with known date and known day of the week. We
select our reference day to be Sunday, January 1,
1978. Since it is a Sunday, we represent it by 0.
(You can select any day you like—the argument will
have to be modified accordingly).
Let X be the number representing the day of the
week on which January 1, 1901 fell. Then we can
write a congruence modulo 7 representing the
number of days between January 1, 1901 and
January 1, 1978.
0 =
+ 77(365) +19-1
Jan 1,
1978
0 =
Therefore
Jan 1,
1901
X + 0
number
of years
•1 + 51
no.
leap
= X +
X = 2 (mod 7).
of
yr
5
(mod 7)
(mod 7)
So, January 1, 1901 fell on a Tuesday.
(d) Hint: Use the statements in parts (b) and (c)
of the exercise.
(e) Hint: Again, use parts (b) and (c).
(f) Hint: How many days are there in a century
from January 1, 1901 to January 1, 2001 ? From
January 1, 2001 to January 1, 2101 ? Etc. Use parts
(a) and (c) of the exercise.
(g) Hint: Use parts (b) and (c). If a year contains
53 Sundays, what is the first day of the year?
CHAPTER S
Exercises
5.2. Hint: Consider congruence modulo 10. (Note
that a number is congruent modulo 10 to its last
digit.)
5.3. Solution: Let n^ = the number, n can be
written in the form n = \0k -h r (where r is the last
digit, 0 < r < 9)
n' = (\Ok-h ry = lOO/j^ + 20kr + r\
lOO/j^ does not affect the last two digits.
20kr does not affect the last digit; it makes an even
contribution to the next to last digit. Therefore, the
parity (evenness or oddness) of the next to last digit

396
HINTS AND SOLUTIONS
of n^ is the same as the parity of the next to last
digit of H.
Consider the possibiHties for H:
02 = 0 = 00,
1^ = 1 = 01,
2^ = 4 = 04,
3^ = 9 = 09,
4^ = 16,
5^ = 25,
6^ = 36,
V = 49,
8- = 64,
9^ = 81.
The only cases in which the next to last digit of H is
odd, are r = 4 and r = 6. In both of these cases, the
last digit of H is 6. Conversely, if the last digit of H
is 6, then r = 4 or 6 and so the next to the last digit
of H is odd.
5.4. Hint: See Exercises 5.2 and 5.3. It is also
helpful to consider what the digital root of a perfect
square could be.
5.6. Solution: If T = the tens digit and U = the
units digit of «, then n = lOT + U. The product of
the digits is TU. Therefore TU divides (lOT + U).
Therefore T divides (10T+ U) and therefore T
divides U (Theorem 4.1, page 105).
Similarly U divides (lOT + U) and therefore U
divides lOT. We make a chart of the possible values
of U and T in which T divides U and U divides
lOT, as follows:
u
T
1
1
2
1
2
2
3
3
4
2
4
4
5
1
5
5
6
3
6
6
7
7
8
4
8
8
9
9
We thus obtain numbers 11, 12, 15, 22, 24, 33,
36, 44, 48, 55, 66, 77, 88, and 99. Of these, only
11, 12, 15, 24, and 36 satisfy the conditions of the
original problem.
5.8. Hint: lOT + U - (lOU + T) = 9(T - U);
^(T + U) = 9(T - U) for each possible value of k,
what can T and U be ?
5.9. (b) Solution: Let the number = lOT + U;
then its reversal = lOU + T.
(lOU + T) - (10T+ U) :
:9U - 9T
9(U - T) :
Therefore, since 9 is a perfect square, U - T must
be a perfect square. Since U — T < 9, we get
U - T = 0, 1 or 4.
We get the following possible values: 11, 12, 15,
22, 23, 26, 33, 34, 37, 44, 45, 48, 55, 56, 59, 66, 67,
77, 78, 88, 89, 99.
5.11. Hint: If a square = 99Ar = 9 • llAr, yfemust be
divisible by 11 and kl 11 must be a perfect square.
Also, if 11 divides (T + U)(T - U), then either 11
divides T - U or it divides T + U.
5.13. Solution: Write the number as lOOH +
lOT + U; its reversal is lOOU + lOT + H.
If H > U, the difference is 100(H - U) +
(U - H).
Since H > U, we write this as 100(H - U - 1) +
90 + (U - H + 10) so that all the terms will be
positive.
Note that the three digits of this number are
H -U - 1,9, U- H + 10;
reversing the digits, y
H - U - 1; and adding
get U - H + 10, 9,
100(H -U-l) + 90 +U-H+10
100(U - H + 10) + 90 + H - U + 1
we get 100 • 9
+ 180 + 9
1089.
5.15. Solution: One way of proceeding is by asking
questions that divide the remaining possibilities in
half. Or we can use the binary system and with
each question can determine another digit in the base
two representation of the number. That is, we may
begin by asking, "Is the number less than 512?"
((lOOOOOOOOO)two).
If the answer is yes, then the base two
representation of the number has fewer than ten digits,
and we next ask, "Is the number less than 256?"
Etc.
If the answer to the question about 512 is no, then
the number has ten digits in its base two
representation, and the leftmost digit is 1. We determine the
second digit in the base two representation by asking
whether the number is smaller than 768
((llOOOOOOOO)two). Each successive question
determines the next digit in the base two representation
of the number. Since there are ten possible digits,
ten questions may be required to pinpoint the
number exactly. As an example, suppose the number
is 379. We then end up asking the following
questions:
Q: Is it less than 512?
A: Yes (there are at most 9 digits).
Q: Is it less than 256?
A: No [the number is (1 )two]-

HINTS AND SOLUTIONS
397
Q: Is it less than 384 (256 + 128)?
A: Yes [the number is (10 )two].
Q: Is it less than 320 (256 + 64)?
A: No [the number is (101 )two].
Q: Is it less than 352 (320 + 32)?
A: No [the number is (1011 )twol-
Q: Is it less than 368 (352 +16)?
A: No [the number is 10111 )two]-
Q: Is it less than 376 (368 + 8)?
A: No [the number is 101111 )two].
Q: Is it less than 380 (376 + 4)?
A: Yes [the number is (1011110—)two].
Q: Is it less than 378 (376 + 2)?
A: No [the number is (10111101-)two].
Q: Is it less than 379 (378 +1)?
A: No [the number is (10111101 l)two = 379].
5.16. Hint: See the solution of Exercise 5.15.
5.18. Hint: Note that card A contains all numbers
up to 31 that have a 1 in the units digit (the last
digit) in their base two representations; card B
contains those numbers that have a 1 in the two's
digit (the next to last digit); etc.
5.19. (a) Solution: On the first deal, all numbers
having 0 as the units digit in their base two
representations are placed, in their original order, on top of
zero, and all numbers having units digit 1 are placed
on one. When the deck is reassembled, it is in the
order shown at the left below.
The cards are now sorted according to their last
digits. The next deal sorts the cards according to the
next to last digit, as at the right below.
—0 —GO —10
—0
— 1
-1
-GO
-Gl
—Gl
-IG
-11
— 11
When the deck is reassembled, we obtain the
situation shown at the left at the top of the next
column.
The next deal sorts according to the third digit
from the right. After reassembly, the result is as
shown at the right.
The final deal completes the sorting, with the
original order restored.
00
Gl
-10
-11
-11
-GOG
-GGl
-GIO
-Oil
-lOG
-IGl
-IIG
-111
OOOG
OGGl
nil
5.20. Hint: At the time that the magician was
blindfolded, how many matchsticks remained on the
table?
If A represents the number of sticks originally
given to the person who ended up with the apple,
B represents the number of sticks originally given
to the person who ended up with the banana, and P
represents the original number of sticks given to the
person who ended up with the peach, what is the
physical significance in this problem of the number
whose base three representation is (PBA)three'^
5.22. (b) Solution: We follow the card's position in
the deck as the magician performs the trick. When the
magician first reassembles the deck, after having dealt
out the three piles, how many cards are above the
selected card? Since each pile contains 9 cards, if the
selected pile was placed on top, then there are
0 • 9 + r cards above it where 0 < r < 8; if the
selected pile is placed in the middle, then there are
1 • 9 -I- r cards above it where 0 < r < 8; if the
selected pile is placed on the bottom, there are
2 • 9 + r cards above it where again 0 < r < 8. That
is, there are 9x + r cards above the selected card
where x = 0, 1, or 2, depending on whether the
selected pile was placed on the top, middle, or
bottom.
When the deck is redealt face up, how many cards
on the table are below the selected card? Since the
9x + r cards that were previously above the selected
card are distributed into three piles, there will be
\(9x + r) = 3x + 5 cards below the selected card.

398
HINTS AND SOLUTIONS
where 0 < 5 < 2 since 0 < r < 8 (there cannot be a
fraction of a card).
When the deck is reassembled, there will be
9y + 3x + s cards above the selected card, where
y = 0,1, or 2, depending on whether the selected pile
was placed on the top, middle, or bottom.
Similarly, when the cards are dealt face up for the
third time into three piles, there will be 3jy -I- x + (0)
cards below the selected card in its pile [note + (0)
because 0 < 5 < 2, and we can't have a fraction of
a card].
When the deck is reassembled this time, there will
be 9z + 3y + X cards above the given card where
2: = 0, 1, or 2 depending on whether the selected pile
is placed on the top, middle, or bottom.
Therefore, if we wish to select the card to be at the
Nth position, we must have
9z + '^y + X = N - \.
That is, (z,yi x) is the base three representation of the
number N - \.
To apply this finding to part (a) of the problem
if N = 23,
N- 1 = 22 = (211)three.
Therefore we place the selected pile first in the
middle (1), next in the middle (1), and lastly on the
bottom (2).
5.24. Solution: In the base three representation of
a number, the digits 0, 1, and 2 are used. In a sense,
though, we could use 0, 1, - 1, since
2 • 3* = (3 - 1)3* = 1 • 3*+i - 1 • 3*.
Thus, any number can also be expressed as a sum
of powers of 3 with coefficients 0, 1, and - 1. For
example,
23 = 2 • 32 + 1 • 3' + 2 • 3"
= 2 • 32 + 1 • 3' + (3 - 1)3«
= 2 • 3^ + 1 • 3' + 1 • 3^ - 1 • 3«
= 2 • 32 + 2 • 3' - 1 • 3"
= 2 • 32 + (3 - 1)3^ -1-3"
= 2 • 3^ + 1 • 32 - 1 • 3' - 1 • 3"
= 3 • 3^ - 1 • 3' - 1 • 30
= 1 • 3^ - 1 • 3^ - 1 • 30
= 1 • 33 + 0 • 32 - 1 • 3^ - 1 • 3".
This representation of a number N is the basis
of this trick. Card A contains those numbers whose
last coefficient (coefficient of 3°) is + or - 1;( - 1 is
indicated by an *); card B contains those numbers
whose next to last coefficient (coefficient of 3') is
+ or - 1 (- 1 is indicated by *), etc.
Therefore, to recapture a number, add the proper
powers of 3 corresponding to cards containing the
selected number without an *, and subtract that
power of 3 if the number on the card has an *.
For example, if N = 8, which appears on Card A
with an * and on card C without *,
add 9 for card C,
subtract 1 for card A,
getting 9-1=8. Note 8 = 1 • 3^ + 0 • 3 - 1 • 3°.
The only problem that can arise occurs when the
result of the calculation is negative. In this case, we
simply add 27 (3^) to this negative number. For
example, 22 appears on cards A and B without *,
and on card C with *.
Add 1 for card A,
add 3 for card B,
subtract 9 for card C,
getting -5. We add 27 to -5, getting 22. (Note
22 = 1 • 3^ - 1 • 32 + 1 • 3' + 1 • 30.)
5.25. Hint: Refer to the solution of Exercise 5.24.
5.28. Hint: U (\\\\\)b = b' -h b' -h b' + b + \ =
b
x^, is X larger or smaller than 6^ + - ?
What about b^ + ? What must x be equal
5.29. Hint: What could the value of 792 -^ 297
possibly be?
5.31. (a) Solution: (121)^ = b^ + 2b + \
= (b+ \y
= [(\\)tV, for each b > 2.
5.32. (a) Solution:
2(b- \) = b-h(b-2) = [\(b-2)]t,
where b - 2 is the units digit and 1 is the b digit.
Similarly,

HINTS AND SOLUTIONS
399
(b- \y = b' - 2b+ \ = b(b - 2) + 1
= [(^-2)l],.
5.35. Hint: Construct a table of squares from 1 to
1000.
5.39. Solution: Since O • T ends in T and N • T
ends in T, the possibilities are:
T = 0,
T = 5 with O and N odd,
or T is even and O and N are 1 and 6 or 6 and 1.
Neither N or O can equal 1 since N • LET ^ LET
and O LET ^ LET. Also T ^ 0 since O + T =
E, hence T = 5 and O and N are odd. Since
N LET = NOT, L = 1.
If E = 0, then O • LET would start with O; but
it doesn't. Therefore E > 2. But 125 • 9 = four
digits, so neither N nor O can be 9. Therefore N
and O are 3 and 7 or 7 and 3.
Since 7 • 145 = four digits, E = 2 or 3. Since
O + T ends in E and O and T are odd, E is even.
Therefore E = 2 and LET = 125.
Since N • LET starts with N and O • LET does
not start with O, N = 3 and O = 7.
Therefore
125
37
~875
375
4625
5.40. Hint: First find C. What is the relationship
between D and E?
5.43. Hint: For reference purposes, represent the
missing digits by letters. (Here, distinct letters might
represent the same digit.)
ab3c
X de f
g hi Oj
kmn7 p
q23 r
s 4 t uv 5j
First find p; then find c and e, then d, then b, and
then a.
5.44. Hint: What could X possibly be? Note that
X^ + the carryover from the previous digit (if there
is any such carryover) = _X.
5.45. Hint: XXX = X • (111) and 111 = 3 • 37.
5.46. Hint: See the hint for Exercise 5.45.
5.47. Hint: How large could CH possibly be? Find
EIN from 7(EIN) = _44_.
5.48. Hint: Is the first digit of the quotient greater
than, equal to, or less than 8?
5.52. Hint: Label the rows of the division problem
as in the solution of Sample Problem 5.4 (page 160).
What could row 1 be? What does this imply about
the divisor?
5.53. Hint: What can be said about the parity of the
last digit of the divisor? What is the relationship
between the last two digits of the quotient?
5.56. Hint: Note that the product of the divisor
and the last digit of the quotient ends in 000.
5.57. Hint: If the square of a three-digit number
has only five digits, what can you say about the
size of the first digit of the three-digit number?
(Remember that you are working in base six.)
5.58. Solution: In whatever base you are working,
ABCABC = (lOOl)ABC.
(1001)seven= 344ten •
(ABC)seven = 49A + 7B + C.
Therefore (344)(49A + 7B + C) is a perfect square,
and so
4(86)(49A + 7B + C) = x^
Since 86 contains no square factors, 49A + 7B +
C = 86 times a perfect square (see Practice
Problems 4.D, problem 5, page 111). Since A, B, and
C are each at most 6, 49A + 7B + C is at most 342.
86 • 2^ = 86 • 4 = 344 is too big.
Therefore, 49A + 7B + C = 86 • 1 = 86. We write
86 in base 7:
86 = 1 • 49 + 5 • 7 + 2
= ( 1 J2)seven >
A = 1, B = 5, and C = 2.

400
HINTS AND SOLUTIONS
5.59. Hint: See the solution of Exercise 5.58.
5.61. Hint: Consider even bases and odd bases
separately.
CHAPTER 6
Exercises
6.3. Hint: Use Theorem 6.5.
6.4. Hint: Use Theorem 6.5.
6.5. Hint: The chessboard may be represented by
a graph whose vertices correspond to cells of the
chessboard. Two vertices are connected by an edge
if it is possible for the knight to move (in one move)
between the corresponding cells of the board.
6.6. Hint: Represent the chessboard by a graph.
A trip is possible for part (a). No such trip is possible
for part (b).
6.7. Solution: There are 6 odd vertices, A, B, C, E,
F, and G. Since we are looking for an Eulerian
path from A to B, we must eliminate " the oddness"
of the vertices C, E, F, and G. We want to do
this in the most efficient way (by adding paths of
shortest possible length).
Clearly, if we add a path from C to F (or retrace
CF) and add a path from G to E (or retrace EG),
we will accomplish this aim. The result is this graph:
An Eulerian path can be found in the manner
described in the chapter. One such path is
ACFDCFGEDGEBDAB. The total distance
traveled with this route is 1400 meters.
6.8. Hint: Refer to the solution of Exercise 6.7,
but note that here not all edges are of the same
length.
6.10. Hint: Consider the graph with 7 vertices
labeled 0, 1,2, 3, 4, 5, 6, with each pair of vertices
connected by an edge. Each edge can be thought of
as representing a domino. For example, the edge
connecting vertex 3 to vertex 6 corresponds to the
domino three-six. With regard to this graph, to what
does a complete chain of dominoes correspond?
What if one domino is missing?
6.11. Solution: Suppose we look at a sequence of
digits—say, 12132—on the circle. This gives rise
to the set of triples 121, 213, 132. Notice that
the pair 21 is the end of 121 and the beginning of
213. Similarly, 13 is the end of 213 and the
beginning of 132. So think of a triple as a merging of
two pairs (for example, 213 represents the merging
of 21 with 13). This suggests that we construct a
directed graph as follows:
Make nine vertices that are labeled by the two-
digit numbers 11, 12, 13,21,22,23,31,32,33. Draw
an edge from vertex ij to vertex km, if j = k (for
example, draw an edge from 12 to 23, 12 to 22, and
12 to 21). Note that each edge corresponds to a
triple. For example, the edge from 12 to 23
corresponds to the triple 123. The resulting graph is
pictured below.
Each vertex has in-degree 3 and out-degree 3.
Therefore there exists an Eulerian dicircuit which
may be found as suggested in the text.
One such dicircuit is 11 -► 11 -► 12 -► 21 -► 11 -►
13 -^31 -^ 12 -^23 -^32 -^22 -^21 -^ 12 -^22 -^
22 -^ 23 -^ 33 -^ 32 -^ 21 -^ 13 -^ 33 -^ 33 -^ 31 -^
13-^32-^23-^31-^ 11. This result is illustrated
at the top of the next column.

HINTS AND SOLUTIONS
401
3 1 1
1.
3.
3 •
3 •
1 •
3
.1
• 3
• 1
3 • . . • 2
2 2 2 1
6.12. Hint: See the solution of Exercise 6.11.
6.13. Hint: Which vertices are of degree 2 ? What
does that tell you?
6.14. Hint: Consider vertices of degree 2.
6.15. Hint: Consider vertices of degree 2.
6.17. Solution: Note that, in this illustration, it is
possible to move from a circled vertex only to an
uncircled one, and vice versa. But there are 12 circled
and only 10 uncircled vertices. So even if a runner
starts at a circled vertex, by the time he has visited
the tenth uncircled vertex, two circled vertices will
not yet have been visited. It is impossible to visit
both of these without revisiting at least one uncircled
vertex. When he tries, he is arrested.
6.18. Solution: Since we are interested in passing
from one state of the filters to another, it seems
reasonable to try to represent the problem graph
theoretically.
We can represent the state (on, off) of the three
filters by a triple of the form (r, b, g), where r = 0
if the red filter is off and 1 if it is on, 6 = 0 if the
blue filter is off and 1 if it is on, and ^ = 0 if the
green filter is off and 1 if it is on. For example,
(1, 1, 0) represents the state where the red and blue
filters are on and the green filter is off.
We construct a graph as shown below, with 8
vertices labeled by the triples representing the
possible states of the filters. Two vertices are
connected by an edge if it is possible to go from the
filter state corresponding to one vertex to the filter
state corresponding to the other by changing one
filter. (This means that the two triples differ in only
one of the three positions.)
(0, 0, 0)
(1,1,1)
(1,0,1)
(1,1,0)
(0, 0, 1)
(0,1,0)
(0, 1, 1)
(1,0,0)
We are looking for a Hamiltonian circuit starting
and ending at (0, 0, 0).
By trial and error, we find
(0,0,0)-(0,0, l)-(0, 1, l)-(0, 1, 0)-(l, 1,0)
-^ (1, 1, 1)-^ (1, 0, 1)-^ (1, 0, 0)-^ (0, 0, 0).
That is, the technician first puts on the green
filter, then the blue, then removes the green, then
adds the red, then adds the green, then removes the
blue, then removes the green, then removes the red.
It is interesting to note that if we considered the
triples as coordinates in the 3-dimension rectangular
coordinate system, then the 3-dimensional graph is
just a cube, as illustrated here.
(0, 0, 1)
(0, 1, 1)
(1,0,1)
(0,1,0)
(1,0,0)
(1,1,0)
The problem reduces to finding a Hamiltonian
circuit on the cube.
6.19. (b) Hint: If the board is colored in the usual
manner, how many white cells are there? How many
black? See the solution of Exercise 6.17.

402
HINTS AND SOLUTIONS
6.20. Hint: Represent the chessboard by a graph
where the cells are represented by the vertices. Two
vertices are connected by an edge if the
corresponding cells of the chessboard are diagonally
adjacent. What is the degree of a vertex
corresponding to a corner cell? Of a vertex corresponding to a
cell on the edge of the board ?
6.21. Hint: Make a graph and consider vertices of
degree 2.
6.22. Hint: Make a graph and consider vertices of
degree 2.
6.23. Hint: Method 1 Make a graph with the ten
vertices corresponding to Arthur and his nine
knights. (It is simplest to arrange the vertices around
a circle.) Two vertices are connected if no statement
in the problem rules out the possibility of the
corresponding people sitting next to each other. Then
find a Hamiltonian circuit.
Method 2 The graph could be obtained by first
making a graph with vertices as defined above and
connecting two vertices corresponding to two people
who cannot sit next to one another in one color, say
red. Once this is done, join each two vertices that
have no common edge by a black edge. This black-
lined graph is called the complement of the first
graph and will be the same as the graph of method 1
above.
6.25. Hint: This problem is essentially the same as
the cannibal-missionary problem. In this problem,
however, there is no restriction on who can row.
They can cross the river in 11 trips (a trip in either
direction is counted as one trip).
6.26. Hint: The crossing can be made with 11 trips
(a trip in either direction is counted as one trip).
6.28. (a) Solution: After each round trip, the
number of people on the far shore can be increased
by at most one. Therefore, if a crossing were possible,
at some point there would have to be three people
on the far shore (five on the original shore). At some
other point, there would have to be four people on
the far shore. By the conditions of the problem, any
time only three people are on the far shore, they
would have to be women. Therefore, to have four
people on the far shore, the fourth also has to be a
woman. Who would row the boat back and how
would the men get across?
(b) Comment: The problem can be solved by
making a graph. To describe each state, we need the
following information: How many of each sex are on
the far shore; on the near shore; on the island. As
a result, there are a large number of states (vertices)
to be considered. So the graph-theoretical approach
seems to be unwieldy. A reasoned trial and error
approach seems simpler.
6.30. Comment: Although this problem can be set
up graph-theoretically, graph theory is not really
needed.
Hint: Although Hugo cannot handle the 60 lb
chest, he is able to pick up the key with his mouth
and drop it out of the window. Also note that it may
be necessary for one who has gone down to go up
again later so that someone else can come down.
6.31. Hint: John and Val may be with Jon's bag
if Jon has both their bags.
6.32. Solution: Let n be the number of books. If
there are more books than there are pages in any
one book, then no book can have more than n - 1
pages. Construct a bipartite graph: One set of n
vertices corresponding to the n books in the library,
the other set of « - 1 vertices corresponding to the
numbers from 1 to « - 1. Draw an edge between
a book vertex and a number vertex, w, if the book
contains m pages. How many edges does the graph
contain? Since there are n books, there are n edges.
But since there are only n — 1 "number vertices,"
if none were of degree greater than 1, then the sum
of the degrees of the number vertices would be at
most n — \. This is impossible; hence, at least one
number vertex is incident to at least two edges—that
is, two books have the same number of pages as this
figure illustrates:
possible
number
oipages
book 1 •-
-♦4
Note that what we have really proved here is an
important mathematical result known as the " Cubby
Hole" or "Pigeon Hole" Principle. It says that if we

HINTS AND SOLUTIONS
403
are placing objects in boxes and there are more
objects than boxes, then one box must receive at
least two objects.
6.34. Hint: Note that the interior intersection
points are not vertices (we call them apparent
vertices), so that we are not really changing the
graph if we unravel it in the following manner. Lift
up vertex 1 and the connecting edges to a position
between 5 and 6, as illustrated.
Move 3 to the left, between 7 and 8, etc. When
you finish unraveling, you eventually obtain this:
6.35. Hint: Replace the checkerboard by a graph
where the vertices correspond to cells and the edges
correspond to knight's moves. Then unravel as in the
hint for Exercise 6.34.
6.36. Hint: Unravel the graph as in the hint for
Exercise 6.34, to the extent possible.
6.37. Hint: Note that the path joining the A's cuts
the board into two pieces (a right section and a left).
Where must the path connecting the A*s go with
respect to the E's, C*s, B's, and D*s?
6.38. Solution: What does this problem have to do
with graphs and planarity ? If the houses and utilities
are considered as vertices, then if it were possible to
make all connections, we would obtain a complete
bipartite graph on two sets of three vertices. We
have seen in the chapter that such a graph cannot
be draNvn in the plane. Therefore, there is no solution
in this case.
6.39. Solution: Consider the first person. Either he
(she) is acquainted with at least three other people
or he (she) is not acquainted with at least three
other people. Consider the case where person # 1
is acquainted with three other people—call them # 2,
#3, and #4. (The other case is analogous.) If any
two of these three people are acquainted, then # 1
and these two people are mutually acquainted.
Otherwise, people #2, #3, and #4 are mutually
unacquainted.
The problem may also be viewed
graph-theoretically as follows. Consider a graph with 6 vertices
(on a hexagon) corresponding to the 6 people.
Connect 2 vertices by a red edge if the
corresponding people are acquainted, and by a black edge if
they are not acquainted. We obtain a two-color
coloring of the complete graph on 6 vertices. The
problem then becomes: Show that the graph
contains a triangle, all of whose edges are the same color.
(Such a triangle is said to be monochromatic.)
This fact gives rise to an interesting game called
Sim, which is presented in the exercises of Chapter
7 (see Exercise 7.68).
6.40. Hint: Construct a complete graph on 5
vertices where the vertices correspond to the 5 men.
Connect 2 vertices with a red edge if the men were
sitting next to one another initially. When the men
resume playing, no one sits next to a person that he
sat next to before; therefore, these edges may be
eliminated. What remains of the complete graph on
5 vertices is called the complement of the graph
(in red) which has been eliminated.
6.41. Hint: See the hint for Exercise 6.40. Find a
Hamiltonian circuit that is a subgraph of the
complement of the graph corresponding to the original
seating arrangement.
6.43. Hint: After three queens are moved, you
must still have one queen in each row and one in
each column. You can therefore proceed by trial,
considering the possibility that the queen in the top
row can be moved. If you move this queen to the
top row left column, then the queen in the fifth row
must be moved. Where can it go? Etc.
Although this problem is not handled graph-

404
HINTS AND SOLUTIONS
theoretically, we include it to introduce two graph-
theoretical concepts:
A set of vertices in a graph are said to be
independent if no two are adjacent to each other
(that is, no one edge directly connects the two of
them).
A set of vertices is said to dominate a graph if
every vertex of the graph is adjacent to a vertex
in the set.
In this problem, if the chessboard is translated
into a graph in the usual manner (where edges
correspond to queen's moves), then the set of vertices
corresponding to the cells where the queens are
currently located forms an independent, dominating
subset of the graph.
6.44. Hint: Eliminate those squares that are
dominated by the four queens.
6.45. (a) Hint: See the hint to Exercise 6.43 and use
the "Cubby Hole" principle (see the solution of
Exercise 6.32).
(b) Hint: Which cell in the fifth row from the top
could possibly be used?
6.46. Hint: Consider the chessboard with its usual
coloring (red, black). If a knight is on a red cell,
what can be said about the cells it attacks?
Don't forget to show that the number you obtain
is indeed maximal. This can be done by showing
that you can pair the cells of the board so that at most
one cell of each pair can be occupied by a knight.
CHAPTER 7
Exercises
7.5. Hint: In this game, a position involves two
pieces of information—how many sticks remain
(denoted by r), and how many sticks were taken
by the opponent on the previous move (denoted by f).
We use the following notation:
(r, t) denotes a position of the game.
For example, (8, 2) means that 8 sticks remain and
2 sticks were taken on the previous move.
(r, 7^ t) means that r sticks remain and that the
number of sticks taken on the previous move was
not t.
(r) means that there are r sticks remaining and
that the number of sticks taken on the previous
move is immaterial.
We can now work backward to find winning and
losing positions. Clearly, (0) is a winning position.
(1, 1) is also winning, since the opponent cannot
move; but (1, ^ I) is losing. We can continue in
this manner displaying our results as in the chart
below.
winning
losing
(0)
(1,1)
(1, 7^ n
(2)
(3,3)
(3,/ 3)
(4,4)
(5,5)
(4, /4) (5, 7^5)
winning
losing
(6,3)
(6, 7^ 3)
(7)
(8)
(9)
(10)
(11,4)
(11,/^ 4)
7.7. Hint: In this game, a position involves two
pieces of information—the number of sticks
remaining, and the parity (oddness or evenness) of the
number of sticks you possess. Work backward.
7.9. Hint: Work backward.
7.10. Hint: Do Exercise 7.9 first.
7.11. Hint: Do Exercise 7.9 first.
7.13. Hint: If, in the winning positions of the
regular version of Exercise 7.12, the number of sticks
in each pile is expressed in the binary system (as in
Nim), what can be said about the column sums?
Can a similar observation be applied to this problem?
7.14. Hint: See the hint to Exercise 7.13.
7.16. Hint for parts (a) and (c): Consider n odd
and even separately, and try to apply a pairing
strategy.
7.17. Hint: What is the connection between this
problem and Exercise 7.16?
7.18. Hint: After how many moves will the game
be over?
7.19. (a) Solution: A position of the game consists
of the number of piles and the number of sticks in
each pile. A position of the game with k piles and a,
sticks in the iih pile will be represented by
(a,, ^2, ..., ak).

HINTS AND SOLUTIONS
405
For example, (10, 3) means 2 piles, one with 10
sticks and the other with 3; and (8) means one pile
with 8 sticks. If it turns out in analyzing the game
that (w) and («) are winning positions, then (w, n)
is a winning position. Similarly, if (w) is winning
and («) is losing, then the position (w, n) is losing.
[If (w) and («) are both losing, we can't immediately
tell whether (w, n) is winning or losing.] (1) and (2)
are winning positions since there is no move for the
opponent. For this reason, piles of 1 and 2 may be
disregarded. We proceed to analyze the game by
working backward.
(3) is a losing position, since the opponent will
leave (1, 2).
(4) is winning, since the opponent will have to
leave (3, 1).
(5) is losing, since the opponent can leave (4, 1).
(6) is losing, since the opponent can leave (4, 2).
(7) is winning, since the opponent must leave
(6, 1), (5, 2), or (4, 3), all of which are losing since
they consist of one winning and one losing pile.
(8) and (9) are losing positions, since the opponent
can leave (7, 1) or (7, 2).
(10) is winning since the opponent must leave
(9, 1), (8, 2), (7, 3), or (6, 4), all of which are losing
since each contains one losing and one winning
subpile.
(11) and (12) are losing since the opponent can
leave (10, 1) or (10, 2), both winning.
(13) is losing; the opponent can win by leaving
(8, 5), provided he or she then plays properly.
(This is not quite obvious and requires further
argument.)
(14) is losing; the opponent can leave (10,4) which
is winning because it is made up of two winning
positions.
Thus, A wins by leaving two piles: one with 10
and one with 4.
7.21. Solution: To put this problem in the
framework of the matchstick games, consider the difference
needed to reach 22. That is, the game starts at 22
and proceeds toward 0. With this in mind, it would
seem from our analysis of Sample Problem 7.1 in
this chapter that the winning positions are the
multiples of 5. However, here a position consists not
only of the total reached but also of what cards
remain. Thus, 5 will be a losing position if no
ace remains and a 4 does, because your opponent
will take the 4 and you will then lose.
On the other hand, don't totally discard your
previous knowledge: 5 is still a winning position if
one of each denomination remains; 10 still wins if
at least two of each denomination remain; 15 still
wins if at least three of each denomination remain.
Thus, A cannot start by taking 4, because B will
reach a winning position (15) by taking 3. Similarly,
A cannot start with 3.
If A starts with 2, B can win by taking 3. To
analyze this case we would have to consider a number
of subcases; but we need not do so because we will
show that A can win by starting with 1.
So assume A starts by taking 1. B cannot take 2 or
else A will take 4 leaving 15 and winning because
at least 3 of each denomination are left.
Similarly, B cannot take 4 or A will take 2 and win.
If B takes 1, A can take 4, leaving 16. B is now
forced to take 1 (otherwise A will be able to leave 10
with two cards of each denomination remaining).
A can now take 4, leaving 11. Again B is forced to
take 1, and again A takes 4, leaving 6. At this point,
B loses no matter what she does. Since no aces remain,
B must leave 2, 3, or 4, after which A can bring the
total to 0.
Thus, B's only reasonable course of action is to take
3 on her first move, leaving 18. At this point A takes 1,
again leaving 17. B cannot take 3 or 4 since A can
leave 10 with at least two cards of each denomination
remaining. Regardless of whether B takes 1 or 2,
A should take 4, leaving either 11 or 12 (see the
figure below).
B,
Az
B,
2/15 11
22-
-21 18-
-17'
1^16 12
From here, B cannot allow A to reach 5, so must
take 1 from 11 or 1 or 2 from 12. Again A takes 4.
This leaves 6 or 7—as illustrated below.
B,
22—^21—^—18—^17

406
HINTS AND SOLUTIONS
If 6 is left, then three of the aces have been used;
and if 7 remains, then all of the aces have been used.
If B now brings the total to less than 5, then A wins
immediately. If B brings the total to 5, then all the
aces are gone, and A can win by taking 4.
7.22. Hint: It is easier to work with the difference
between 26 and the previous total. A position
includes two pieces of information: the difference
between 26 and the total; and the number showing
on the top face of the die.
Work backward as in previous problems. Note that
a pattern is established once six consecutive entries
exactly match six earlier consecutive entries.
7.28. Hint: Consider the triple (w, «, p) where m
represents the number of spaces between A and B
on the top row, n on the middle row, and p on the
bottom row. In this light, the game is related to
Nim.
7.29. Solution: (a) By symmetry, there are three
possible opening moves for A: center, corner, or
middle of a side. Call them cases I, II, and III. The
first step in attacking this problem is to play the game
a few times to get a feel for it. Do you get the feeling
that starting at the corner or middle of the side seems
to result in a large variety of possible moves to be
considered, whereas starting at the center
immediately limits the possibilities that must be considered?
Since we are just interested in finding a winning
strategy for A (if one exists), let us consider the
simplest case first.
Case I A starts by placing an X in the center.
B is now forced to place an O, otherwise A will win
on the next move. By symmetry there are only two
possibilities—O in a corner or the center of a side.
In either case, A has several alternatives. To keep
the analysis simple, consider moves that limit B's
possibilities as much as possible. (If this doesn't lead
to a winning strategy for A, we will have to try the
other alternatives.)
Case la In the case that B's first move is the corner,
a quick check of possible moves for A reveals one
that leaves B no safe reply—A places an O diagonally
opposite from B's O, as the figure at the top of the
next column shows.
Case lb If B's first move is the middle of the side,
then the move for A that most limits B's moves is to
place an O symmetrically opposite to the O on the
board, as in the figure below. B now, by symmetry.
has only one move—an O in one of the remaining side
boxes. A now forces a win by filling in the fourth side
box with an O, as in the figure below:
O
Thus, A has a winning strategy: Place an X in the
middle box and play O's symmetrically opposite to
B's, until B presents A with an opportunity to win.
Note that, since case I has led to a winning strategy
for A, it is no longer necessary for us to consider cases
II and III. On the other hand, if some possible
continuations in case I would have led to a win for
B, it would have been necessary for us to consider case
II (A starts in a corner) and maybe even case III
(A starts in the middle of a side).
(b) The misere form of this game is much harder
to analyze. A can ensure at least a draw by starting
in the center and then playing symmetrically opposite
to B by using the symbol not used by B. An example
is shown below:

HINTS AND SOLUTIONS
407
X
X
If A's move would complete three in a row, then
B would have already completed three in a row on her
previous move. On the other hand, the question of
whether or not A has a winning strategy is more
difficult. In order for A to force a win, he must leave
a position such that when B moves she will be forced
to make three in a row with either an O or an X.
Since B makes the second, fourth, and sixth moves in
the game, the only possible positions in which B will
be forced to complete three in a row are those shown
in the figure below, and their symmetric counterparts
(including switching X's and O's).
o
X
X
X
X
o
o
X
o
X
o
o
X
X
o
o
X
o
X
X
X
(a)
(b)
(c)
o
o
X
X
o
X
X
o
o
X
X
o
X
X
X
o
X
X
X
o
o
(d)
(e)
(f)
Can A always leave one of these positions, or can
B prevent this from happening? B can prevent it, but
to see this, many cases must be considered. Let us
consider just one example here: If A starts by placing
an X in the center, then B can place an O on the side.
(Only positions (c), (d), and (e) are now possible.)
And if A continues by placing an O in the corner,
then B can place an X in an adjacent corner, as
illustrated here. No matter what happens at this
point, we cannot end up with any of the above end
positions.
7.30. Solution: We use the following notation:
X, indicates A's ith move and O, indicates B's ith
move. There are two possible opening moves for
A—bottom corner or bottom middle.
(a) Case I A opens in a bottom corner. B has
three possible responses, illustrated here:
lb
Ic
Case la Part a of the figure below shows that A can
force a win. (Note that O2 and O^ are forced moves,
and that O3 is essentially forced, otherwise A will win
immediately thereafter.) Since case la doesn't lead to
a win for B, we must consider case lb.
02
X
;^.1
0,
/ ^
0,
/^5
O4
X4
0,
0,
/^
X4
/^
O2
/ 5
O4
X2
la
lb
Case lb Again A can force a win, as in figure b
above. Here all of B*s moves are forced. We now try
case Ic.
Case Ic A can force at least a draw by playing above
B's move and then leaving one of the positions below.
X3
02
X,
x^
0,
X,
o,
On the other hand, no matter what A does, B can
play directly above A's move and continue to do so
until B gets the center square (this must happen
eventually). With best possible play by A (you supply
the details), we get the figure shown below, where A
can place his X above the O and the game is a draw.

408
HINTS AND SOLUTIONS
Thus, case I results in a draw if both players play
well.
We now must consider case II.
Case II A starts in the bottom middle. B has two
possible responses, as this figure shows:
O,
Ila
Xi
lib
Case Ila is a theoretical draw. An example is
illustrated below.
03
X3
02
X2
Oi
X,
X5
04
X4
o,
X3
Case lib leads to a win for A, as shown below.
X4
0,
/^
Oz
7^2
X,
/ 5
O4
0.
Note that B's moves are essentially forced. Thus,
case II also leads to a draw since B can respond as
in case Ila.
Since both cases lead to a draw with proper play,
the game is a theoretical draw (both players have
drawing strategies but neither has a winning
strategy).
(b) In the misere form of this game, A wins by
forcing B to complete the middle row. That is, A
starts anywhere, and whenever B plays the middle
row, A plays on top of B; if B plays on the bottom
row, A takes the remaining space on the bottom row.
7.37. Hint: By symmetry, the possible first moves
for A are as in the figure in the next column.
In all but one of these cases, B has a move to
force a win. In the one remaining case, A has a
winning strategy.
X,
0,
0,
X.
(a)
(b)
0,
X,
X,
0,
(c)
(d)
0.
X.
0,
X.
(e)
(0
7.39. Hint: This game is different from Sample
Problem 7.2 in the chapter in that the X's (checkers)
must be in touching boxes. As a result, not all
positions that were equivalent before are equivalent
now.
7.40. (b) Solution: A can ensure a draw by starting
in the center and then using central symmetry to B's
moves. (For example, if B moves in the upper right-
hand comer of the top level, A moves diametrically
opposite in the lower lefthand comer of the bottom
level. Etc.) Thus, A has a drawing strategy. [See the
solution of Exercise 7.29(b).]
However, the game cannot end in a draw. To see
this in the case where A adopts the above strategy,
first observe that there are essentially four different
types of cells in a 3 x 3 x 3 Tic Tac Toe cube. These
are illustrated on page 369. There is the middle cell
(labeled M), the eight corner cells (labeled C), the
twelve non-corner cells that are on an edge of the cube
(labeled E), and the six cells that are centers of faces
of the cube (labeled F).

HINTS AND SOLUTIONS
409
.^^7^^
.//7
C I
I F
v\
X
o
X
O
O
X
X
X
o
bottom
level
X
?
middle
level
X
O
O
O
X
X
top
leve
O
X
O
1
M = cell in the center
c
E
C
E
F
E
C
E
C
E
F
E
F
M
F
E
F
E
C
E
C
E
F
E
C
E
C
bottom
level
middle
level
top
level
Assume that A has adopted the indicated strategy
and that the game has ended in a draw with all the
cells filled in. Then, by the nature of the strategy,
there must be four X's and four O's in the corner
cells (C), because each time B took a corner cell, A
took one too.
If all four of the corner X's appear in the same
horizontal, vertical, or diagonal plane, then some
portion of the cube looks like the figure below.
X
X
X
X
There is no way that this plane can be completed
without someone being a winner.
If three of these corner X's appear in the same
horizontal or vertical plane, then by rotating the cube
if necessary, there is no loss in generality in assuming
that the plane in question is the bottom plane and
that the position is as shown here:
X
X
X
o
X
X
o
o
o
bottom
level
middle
level
top
level
Since the position is supposedly drawn, there is only
one way in which the bottom and top levels can be
completed:
But now there is no way to complete the middle level;
the cell with the question mark cannot receive an X
or an O without one player completing three in a row,
as this figure shows:
Hence, this case leads to a contradiction.
The only remaining possibility for the four corner
X's is as shown below (or an equivalent position).
o
X
?
X
o
X
X
o
o
X
bottom
level
middle
level
top
level
But now there is no way to fill in the cell that has the
question mark. Hence, all cases lead to contradictions
and so, if the first player follows the indicated
strategy, the game cannot end in a draw. But as the
indicated strategy is a drawing strategy (ensures at
least a draw), the first player cannot lose and therefore
must win by following this strategy.
7.41. (b) Hint: See the solution of Exercise 7.30(b).
7.50. (a) Hint: From a position such as that in
part a of the figure below, A cannot be cut off from
the top of the board. Similarly, from a position such
as that in part b, B is forced to move to the cell
indicated by the arrow, if she wants to cut A off from
the top of the board. From a position such as that in
part c of the figure, A cannot be cut off from the top

410
HINTS AND SOLUTIONS
Black
Black
Black
of the board, since if B moves to the cell indicated
by the arrow, then A can move to the cell indicated
by *, ensuring a connection with the top of the board.
(b) (i) Solution: B wins by labeling the cells as
shown in the figure below and then playing as
follows: Whenever A moves to a lettered box, B
moves to the other box with the same letter. If A
moves to the center, B moves to any box and then
plays the pairing strategy until unable to do so, in
which case she makes another free move. Etc.
Black
White
At the end of the game, by following this strategy,
B will never have a center cell, more than one W cell,
more than one X cell, more than one Y cell, or more
than one Z cell. So B cannot have a path connecting
two sides of the board. Since the game cannot end in a
draw, A must have a path connecting the top to the
bottom of the board. So B wins.
(b) (ii) Solution: In the 4x4 case, A can win by
labeling the board as in this figure and by playing as
Black
White
follows: He starts with H and adopts the pairing
strategy discussed above for (b) (i). He never plays in
the unlabeled cell. A also never occupies both cells
with the same letter, so there is no way for A to have
a path from the bottom to the top of the board.
(b) (iii) and (iv) Comment: It is known that, for
general odd n, B wins the n x n game and, for general
even «, A wins the n x n game. We do not know the
winning strategy.
7.52. Hint: If at the end of the game, an odd
number of dots has been used, then an even number
of line segments has been drawn, and B made the
last move. The reverse is true if an even number of
dots has been used.
We are not certain of the general strategy of this
game; however, it seems to us that the dot counting
idea mentioned above should be applicable.

HINTS AND SOLUTIONS
411
7.54. Hint: (i) Use symmetry.
(ii) For each possible opening move by A, show
that B has a successful countermove.
7.56. Hint: Consider how many moves A should try
to make in order to wm the game. Note that A
may be able lo influence the number of moves in
the game by choosing among patterns such as those
shown in the figure below.
r I I I I 1
Also consider the 2x4 game and note that there
is a relationship between what happens in the 2 x «
game and what happens in the 2 x (w + 4) game for
most values of n.
7.57. (c) (i) Solution: The method is to take cases
for A's first move. By symmetry, A has four possible
opening moves. Rather than show the complete
analyses, we take one opening move that does in fact
lead to a win for A.
Let A start as in the figure below. Then, by
symmetry, B has four possible responses. These are
shown in parts a , b, c, and d of the next figure.
(c) (d)
In each case, A can reply as shown below:
(a)
(b)
(c)
(d)
In each case, no matter what move B makes, A at
his next turn can eliminate one of B's possible moves,
and so will win the game.
(c) (ii) Solution: Again we consider a winning
opening move for A, as shown in the first figure
below.
If B doesn't prevent A from moving as in parts a
or b of the next figure, then the game will end after
r^T
(a)
(b)
(a)
(b)

412
HINTS AND SOLUTIONS
B*s fourth move and B will lose. Therefore, B has only
two reasonable moves, as shown below.
(a) (b)
In these cases, A responds as follows:
(a) (b)
In both of these cases, B will have to make the
last move.
7.61. (b) Hint: We present a partial solution. We
have found a winning strategy for A which starts with
the position in the first figure below.
B
A
B
A
B
A
B
A
To this move, B has four possible responses, as
shown in parts a, b, c, and d below.
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
(a)
(b)
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
In a complete solution, all of these cases must be
considered. Suppose B moves as in part a of the
figure, above. A's best response is shown in the single
figure below.
B
A
B
A
B
A
B
A
Again B has several alternatives. These are shown
here as a, b, and c.
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
(a)
(b)
B
A
B
A
B
A
B
A
(c)
We will consider only what happens in the a case.
Again, you should consider all the other cases. (You
can see that to complete this problem, an extensive
case analysis is necessary.)
If B moves as in part a of the figure, A can now win
by moving as below (sacrificing the piece on the
right):
B
A
B
A
B
A
B
A
(c)
(d)
B is forced to move the threatened pawn in the
third column from the left (either to capture or to
push), otherwise A will win on the next move. But
either way, A can capture B's pawn in the second
column from the left and win on the next move.

HINTS AND SOLUTIONS
413
7.66. Hint: Does the size of the table matter?
Does symmetry play a role?
7.67. Hint: In order to cut down the amount of
work needed to analyze this game, we must consider
what positions are equivalent to each other.
Consider the positions in parts a and b of the
figure below. Are they really different from each
•2
(a)
(b)
1
(c)
Other? It is true that in part a, points 1 and 2 are
inside the loop, whereas in b they are both outside.
But any line that can be drawn in part a has an
analogous line that can be drawn in part b. On the
other hand, the position in part c of the figure is
different from those in a and b. In c, it is not
possible to draw a line connecting 1 and 2 whereas
it is possible in the other figure parts. Thus, in
considering the game of Sprouts, the notions of inside
and outside are not as important as whether or not
points are separated from one another.
Therefore, in the 2-dot game, there are (by
symmetry) only two starting moves that the first
player can make—see parts a and b below. Note
that the positions in parts b and c of the figure
are equivalent.
1^^^2
Ol -2
'GD
(a)
(b)
(c)
Another consideration in analyzing this game is:
How long can the game last? (What is the maximum
length of the game?) At the start of the game there
are n dots, each having three lives. With each move
of the game, two lives of existing dots are eliminated
and a new dot having one remaining life is added:
There is a net loss of one life. If at any time there
is only one life remaining, then it is not possible
to make a move and the game is over. Thus, the
game can last at most 3« - 1 moves (starting with 3«
lives). Note that it is possible for the game to end
in fewer than 3« - 1 moves if we are left with
several dots, each having one life, that are separated
from one another. This fact influences the strategies
of the two players. For example, the 2-dot game
would last at most 3-2-1 = 5 moves. Thus, A
would make the last move if all 5 moves were
available. Therefore, in this game, B would try to
isolate a dot, so that the game will end in four moves.
A tries to prevent this.
7.68. Hint: (a) (i) For w = 5, there are 10 lines
that could be drawn; A will draw 5 of these and B
the other 5. If A can complete a circuit (see Chapter
6) through 4 of the dots and connect the fifth dot to
one of the four (as in the figure below, for example),
then B will have to take the remaining five lines and
will lose. A*s strategy is to try to complete such a
figure and B*s strategy is to prevent A from doing so.
(a)
(b)
(b) Hint: See Exercise 6.39 in Chapter 6.
Comment: The 6-dot version of Sim is the one
most frequently played. It is still of interest even
though it has been shown by exhaustive computer
searches that the second player has a winning
strategy. L. Shader ([41], Vol. 51, pp. 60-64) has
also found a winning strategy for the second player
without the use of a computer.
7.70. Hint: To help in the analysis, label the cells
with numbers as shown in this figure:
7
14
21
8
15
22
1
4
9
16
23
28
31
2
5
10
17
24
29
32
3
6
11
18
25
30
33
12
19
26
13
20
27

414
HINTS AND SOLUTIONS
Unless the fox is able to jump a goose, after an
odd number of moves by the fox he will be on an
even numbered cell; and after an even number of
moves, the fox will be on an odd numbered cell.
7.71. Hint:
positions.
Work backward to find winning
7.72. Hint: Tempo (timing) is important. Oblique
Street can allow Fred to attain the correct tempo.
7.76. Hint: You should win 6-3 with proper play.
CHAPTER 8
Exercises
8.2. Hint: Let M„ = the number of moves required
if there are n switches in the row and let G„ be
the number of moves required to convert n switches
from the state in which they are all off to the state
in which the «th switch is on and the rest are off.
Express G„ in terms of G„ _ i; find a formula for G„
as a function of n; find a formula for M„ in terms
of M„ _ 2 and G„ _ i. Note that it requires the same
number of moves to turn n "on" switches off as it
does to turn n "off" switches on.
Comment: This problem is essentially the same as
the well-known Chinese Rings Problem (see [3],
p. 305).
8.5. (a) Hint: What would the area of the required
square be? How big would one of the sides have to
be? How could you easily construct a line segment
of the required length?
(b) Hint: What could be the dimensions of the
triangle?
(c) (i) Hint: Label the regions as in this figure:
Which regions make up the square of side c? Are
certain regions congruent?
(ii) Hint: Label the regions as in the figure below.
Show that the four quadrilaterals made up of regions
1, 2, and 3, regions 4, 5, 6, and 7, regions 5, 8, and
10, and regions 6, 9, and 11 are all congruent to
each other.
8.6. Hint: Figure out how many cells have to be in
each section. Then try systematically to describe all
possible sections that could have this number of cells.
For example, in the 4 x 4 case—part (b) of the
exercise—each section must contain eight cells.
There are essentially six different ways of cutting
the board up into identical eight cell sections, as
the figure below shows:
(1)
(2)
1
1
(3)
1
1
i
(4)
(5)
(6)

HINTS AND SOLUTIONS
415
Each of these may give rise to several different
cuttings under rotation or reflection, depending on
whether condition (i), (ii), or (iii) is being considered.
For example, if coloring counts, then the two
cuttings in the figure below are not the same, whereas
they are the same if coloring does not count.
^^
Similarly, if we are allowed to flip pieces over
(without regard to color) then the two cuttings that
follow are the same, whereas if we are not allowed
to flip pieces over, they are not the same.
Thus, for each of the six cuts in the first figure,
we must count how many diff'erent cuts there are
depending on (i), (ii), or (iii). For example, in (i),
cut (2) in the figure gives rise to the four different
solutions shown below.
r
z\
L
In (iii), all four of the cuttings are the same,
giving one solution arising from this type of cut.
8.9. (a) Hint: If the missing cells are in the same
row or column, how far apart can they be? If they are
in diff'erent rows and columns, show that the problem
can be reduced to that of covering an odd-by-even
board that has diagonally opposite missing corner
cells.
(b) Hint: Does the color of the missing cell
matter?
(c) Hint: How many cells of each color can each
of the teirominoes have?
(d) Solution: It has become traditional to refer to
each of the pentominoes by the letter that it most
resembles. These are shown below.
J
L
(1)
(2)
L
(3)
(4)
In (ii), cuts (1) and (4) here are the same as are (2)
and (3), giving two solutions arising from this type
of cut.
W
X Y Z
(f) Hint: Use F, N, P, U, V, and X to make one
rectangle and the others to make the other.

416
HINTS AND SOLUTIONS
(g) Hint: For each pentomino, place it in an
allowable position—a position in which no fewer than
5 cells are isolated—so that the largest possible
number of cells of the pentomino lie on the border
of the given region. In this manner, obtain a
"maximal edge number'' for each pentomino. Add
these up. What do you observe?
8.11. (a) Hint: Where was the remaining center
peg, one move before the last ? Where was it one move
before that?
(b), (c), (d), (h), (i), (k), (1), (m) Hint: Use coloring
arguments.
8.13. (b) Hint: Can this be reduced to the previous
case?
(c) Hint: Can this be reduced to previous cases?
8.14. Hint: Note that there are two T's and two O's.
8.17. Hint: We will refer to the cubes in the
2x2x2 model as UFR (for the cube in upper
front on the right), UFL (upper front left), UBR
(upper back right), etc. Note that L is for left and
L is also for lower. Consider UFL in the first
figure below. It must have blue on top, green on
the left, and red on the front. Let t, x, and z
represent the remaining faces of this cube, as
indicated in the second figure.
UBL UBR
V X V
/
y
/
/
/
UFL
XUFR /
LFL LFR
blue
/
green
Now consider UFR. It must have blue on top,
red on the front, and orange on the right. It also
must have x on the left since it is next to UFL.
Label the remaining faces u and v as in the figure
below. Continue in this manner for each of the other
cubes, introducing a new letter each time an
unknown face appears but being sure to use the same
letter when two faces touch. Note that since x is on
UFL, it cannot be blue, red, or green; and, since it
is also on UFR, it cannot be orange either. Therefore
X must be white or yellow.
blue V
-orange
In a similar manner, you will be left with two
possibilities for each of the other unknowns. If you
choose X to be white, each of the other unknowns
will be determined; if you choose x to be yellow,
then each of the other unknowns will again be
determined. You thus get two solutions, although it
turns out that both solutions make use of the same
eight cubes.
8.18. Hint: When a knight makes a move from a
cell of one color, what can be said about the color on
which he lands? How many moves must he make in
all for a reentrant tour?
CHAPTER 9
9.7. Solution: Label the four coins A, B, C, and D.
Label the fifth coin G.
First weighing: weigh A, G against B, C.
If they balance, write A + G = B + C.
If A, G is heavier, write A + G > B + C. Etc.
The first weighing gives three cases:
Case I A + G = B + C.
In this case, D must be the bad coin. A second
weighing of D against G tells you whether D is heavy
or light.

HINTS AND SOLUTIONS
417
Case II A + G > B + C.
In this case, D is a good coin, and either A is
heavy or B or C is light. A second weighing of
B against C determines which coin is bad. That is,
if B = C, then A is bad and heavy.
if B < C, then B is bad and light.
if B > C, then C is bad and light.
Case III A + G < B + C.
This is essentially the same as case II-
interchange the words heavy and light.
-just
9.8. Hint: Weigh four against four. If they balance,
use the solution of Exercise 9.7. If they don't balance,
regroup and weigh some of these against others for
the second and third weighings.
9.9. Hint: Label the coins A, B, C, D, and E.
First weighing: A against B.
Second weighing: C against D.
By symmetry, suppose A < B and C < D
(otherwise rename so that the first coin weighed in each
pair is the lighter coin).
Third weighing: A against C. (Assume A < C;
there is no loss of generality in doing this.)
Fourth weighing: E against C.
You now have a partial ordering. In one more
weighing you can order four of the coins. Then work
with the fifth.
9.10. Hint: To begin with, there are 120 different
possibilities for which coin weighs what amount.
If you weigh two coins against each other, there are
two possible outcomes, so the best you can do is to
eliminatehalf of the possibilities with each outcome.
However, if you weigh two coins against two other
coins, there are three possible outcomes ( = , <, >).
This may enable you to eliminate two-thirds of the
cases with each weighing.
You must carefully choose which pair to weigh
against which other pair, so that each of the possible
outcomes leaves essentially one-third of the cases.
9.11. (b) Hint: There are 16 oz in 1 lb.
9.12. (b) Solution: In order to ensure that D ends
up facing its original direction, D leaves E and F and
goes via 3 to 1; then leaves 1 via 2 to pick up C (so
far, one reversal).
D pulls C to 2, leaves it there; passes through
1 to 3 then to 4 to pick up A, B (three more reversals).
D pulls A, B to the right and pushes them to 1 and
3 respectively (two reversals). D backs up (very
slightly) to pick up E and F; pulls E, F to the left and
backs up, backing C into A, then moves to the left,
pulling C, A along (four reversals). D pushes E, F, C,
and A to the right on the main track, then heads up
3, pushing B to 1 (two reversals). D then backs up,
pulls E, F, C, A to the left and then backs them up
to pick up B (three reversals). Finally D pulls E, F, C,
A, B to the main track (one reversal), leaves C, A,
B, and continues with E and F on its way.
If Connie is willing to leave the stalled cars on the
triangular tracks, then only eight reversals are
necessary.
9.14. Hint: What positions near the end could lead
to the desired position?
9.16. Solution: A neat method of handling this
type of problem—one involving the expressions "All
A's are B's ", " Some A*s are B*s **, " No A's are B*s '*,
and "x is an A'*—is the method of Euler circles.
We denote the set of all A*s pictorially as the
region bounded by circle A. Similarly, the region
bounded by circle B denotes the set of all B*s. We
can then translate the above statements into Euler
diagrams as follows:
No A's are B's
X is an A
If we now make all composite Euler diagrams
which satisfy the premises of a given argument, we
can check to see if the conclusion holds in every
instance.
In this exercise,
1. All gangs are gengs.
2. Some gengs are gings.
3. No gengs are gongs.

418
HINTS AND SOLUTIONS
The only possible Euler diagram for premises 1
and 3 is shown below.
gengs
gongs
may not be capable of reproducing by
parthenogenesis. Hence, we cannot make the desired
conclusion.
9.24. Hint: Let W„ = the number of worms that
there were at the end of the «th month. What is the
relation between W„, W„ _ i, and W„ _ 2?
The sequence of numbers that we get is called the
Fibonacci sequence.
Since this diagram implies conclusion (a)
regardless of premise 2, conclusion (a) must hold.
However, the following diagrams also satisfy
premise 2 together with 1 and 3.
gengs
gongs
Case I
Conclusion (b) does not hold in this case.
gengs
gongs
Case II
Conclusions (c) and (d) do not hold in this case.
Therefore, conclusions (b), (c), and (d) do not
follow from the premises.
To use this method in general in order to show
that a conclusion is valid, we must show that the
conclusion is satisfied in every possible Euler diagram
which is consistent with the premises. To show that
a conclusion is not valid, we must find one Euler
diagram which is consistent with the premises but
in which the conclusion does not hold.
9.19. Solution: From statements 1, 2, and 4, all
bees are capable of reproducing by parthenogenesis.
From statements 3 and 5, all bees are beneficial to
the farmer.
But there may be insects other than bees that
are also beneficial to the farmer, and these insects
APPENDIX B
B.l. Solution: Let
S„:l+2 + 3 + --- + «:
1(2)
n(n + 1)
S^: 1 = = 1, so S, is true.
Assume S* is true.
1 + 2 + ••• +/? =
k(k-h I) k" -h k
Add /? + 1 to both sides.
k^ + k
k'
k'
(k
+ k-h 2k -{- 2
2
+ 3k + 2
2
+ \Xk + 2)
1 + 2 + • • • + (yfe + 1) =
z
This last equation is just S^t + i-
Therefore, S^ implies S^ + , and so, by induction,
since S^ is true, S„ is true for all « > 1.
B.3. Solution: Let
S„: Any map formed by n straight line segments
(each beginning on one edge of the rectangle and
ending on another edge) can be colored in two colors
so that bordering regions are of different colors.
Then
S,: Any map formed by 1 straight line segment
in the manner defined above can be colored in two
colors.

HINTS AND SOLUTIONS
419
Sj is clearly true since there are only two regions,
as the figure below exemplifies.
Assume Sk is true. We wish to show that S^t 4-1 is
true. We illustrate our argument with figures for
k = 4. Suppose we have a map formed within a
rectangle by /? + 1 straight lines in the manner
described above. (For example, see the figure below.)
jfe = 4, k ^ \ = 5
We must show that it can be colored in two
colors. To do this, we delete any one of the k + 1
lines. This leaves a /f-line map.
The dotted line is the one being deleted.
Since S* is assumed to be true, this map may be
colored in two colors.
Now reinsert the line that was previously deleted.
Reverse all the colors on one side of this line,
leaving the colors on the other side of the line
unchanged.
We now have colored the original map in two
colors so that bordering regions are in different
colors.
To check this, suppose that we have two bordering
regions. Their common border is either part of one
of the k lines or part of the /? -I- 1st line. In the
former case, the two regions lie on the same side
of the /j + 1st line, and therefore either remained
unchanged in color or both were reversed in the final
coloring. Since they received different colors in the
/j-line coloring, they are still of different color in the
final coloring.

420
HINTS AND SOLUTIONS
In the case that the common border is part of the
yfe + 1st line, the two regions were part of the same
region in the /j-line coloring, and only one of them
was changed when the /? + 1st line was reinserted.
They therefore have different colors in the final
coloring.
Thus, S;t + 1 is true. We have just shown that if Sk
is true, then S^t + i is true. Therefore, by mathematical
induction, since S^ is true, S„ is true for all « > 1.

HINTS AND SOLUTIONS
421
Appendix C Hints and Solutions
C.i Hint: What could the sum of each
pile possibly be?
C.2 Hint: Break this problem up into
cases. This may be done by counting all the
triangles of size i, all made up of four basic
triangles, etc.
C.7 Hint: How many arrangements of
five names are there? How many of these
are in alphabetical order?
C.io Solution: Suppose red is one of the six
colors. Then one face must be red. Turn the
cube so that the red face is on top. This
leaves five possibilities for the color of the
bottom face. Once the color of the bottom
face is chosen, there are four colors
remaining. Say blue is one of these colors,
without disturbing the top and the bottom
faces of the cube we may rotate the cube so
that the blue face is now in front. There are
now three possibilities for the color of the
back face (opposite to the blue one); that
leaves two possibilities for the color of the
face on the right. See figure 46.
top
left-
back
right
\
bottom
FIGURE 46
The color of the face on the left is then
forced (only one left).
Hence by the Multiplication Principle,
the are 5-23 = 30 possible colorings.
C.13 Hint: In how many ways can the
tablecloths be placed on the top row? Once
this row is placed, in how many ways can the
second row be placed? What about the third
row?
C.14 Hint: Once the occupant of Dioi is
known, how many possibilities are there for
D102? Continue in this manner but be
careful.
C.15 Hint: We are essentially asking the
follovdng question. In how many ways can
we fill in the array in Figure 47 with the
names Manny, Brooke, William and Queenie
so that no names appear more than once in
any row or any column?
North
South
East
West
Night
1
night
2
night
3
night
4
FIGURE 47
How many ways are there to fill in the
North row? Once the North row is filled in,
how many ways of filling in the East row
depends on how the North and South rows
are filled in.
Therefore we must subdivide the seating
arrangements of the South row into two
cases.
C.17 Partial Solution: Break the counting
procedure up into disjoint cases:
Case 1: No color is repeated - (This
case has been handled in Problem C.io).
Case 2: One color appears twice and
the remaining colors are different.
Case 2a: The repeated color occurs
on opposite faces.
Case 2b: The
on adjacent faces.
Case 3: One color appears three
times and the remaining colors are different.
Case 3a: Two of the faces of the
repeated color are opposite one another.
Case 3b: No two of the faces of the
repeated color are opposite one another.
Case 4: One color appears four
times and the remaining colors are distinct.
Case 4a: The repeated color appears
on two pairs of opposite faces.
Case 4b: The repeated color does not
appear on two pairs of opposite faces.
Case 5: One color appears five times
and the other color is different.
Case 6: One color appears six
times.
Case 7: Two colors each appear
twice and the remaining two colors are
different.
repeated color occurs

422
HINTS AND SOLUTIONS
Case 7a: Each repeated color appears
on two opposite faces.
Case 7b: One repeated color appears
on opposite faces but the other does not.
Case 7c: Neither repeated color
appears on opposite faces. But the
remaining two colors are opposite each
other.
Case 7d: Otherwise.
Case 8: Three colors each appear
twice.
Case 8a: All pairs of repeated colors
appear on opposite faces.
Case 8b: One repeated color appears
on a pair of opposite faces, but the others do
not.
Case 8c: No repeated color appears
on a pair of opposite faces.
Case 9: One color appears three
times, another appears twice, the remaining
color is different.
Case 9a: The doubly repeated color
appears on opposite faces.
Case 9b: The triply repeated color
appears on a pair of opposite faces, but the
doubly repeated color does not.
Case 9c: Otherwise.
Case 10: Two colors each appear
three times.
Case 10a: A pair of opposite faces is
the same color.
Case 10b: No pair of opposite faces
are the same color.
Case 11: One color appears four
times and another color appears twice.
Case 11a: The color which appears
twice is on opposite faces.
Case lib: The color which appears
twice does not appear on opposite faces.
In each case, first count the number of
ways in which the colors can be chosen, then
count the number of ways in which the
chosen colors can be arranged and then use
the Multiplication Principle.
For example, in case 7a there
are ways of choosing the two repeated
colors and then there are ways of
selecting the remaining two colors. Once the
colors are chosen, there is only one way of
arranging them. That is, suppose red and
black are the two repeated colors in 7a we
can turn the cube so that red is on the front
and back faces, black is on the left and right
faces. Whichever of the remaining colors is
on top is immaterial because the cube can be
turned over. Therefore, case 7a gives rise to
= 90 possibilities.
Now in case 7b the analysis is a little
different, as the two repeated colors are not
equivalent. Namely there are 6 possibilities
for the repeated color to appear on opposite
faces, this leaves 5 possibilities for the other
possibilities for the
remaining colors. Again there is only one
way to arrange these colors. (For example in
Figure 48, cube B is obtained from cube A by
svdtching front and back and rotating by
180° about an axis from front to back.)
^4^
repeated color, and
This gives 6 5
2
^
1
180 possibilities.
2
T
FIGURE 48
In 7c there are | " ways of selecting the
repeated colors. There are ways of
choosing the remaining colors. Again there
is only one way to arrange them.
^6Y4^
So there are
2A 2
90 possibilities.
In case 7d again there are
ways
of choosing the colors, but this time once the
colors are chosen there is more than one way
of arranging them since we lack the
symmetry of the other cases. Suppose the
repeated colors are red and black and the
other two colors are blue and green. Since
the red faces are not opposite each other, a
red must be opposite a black. We may turn
the cube so that the red is in front and black
is in back. Without loss of generality put the
other red face on top. Since the blue face is
not opposite the green face, the remaining

HINTS AND SOLUTIONS
423
black face cannot be on the bottom. So there
are two possibilities (right or left) for the
position of the remaining black face. Once
this position is determined there are two
possibilities for the placement of the blue
and green faces. Therefore there are
■■ 90 ways of choosing the colors;
and there are 2 2 = 4 ways of arranging the
chosen colors on the cube. Thus case yd
gives rise to 90 • 4 = 360 possibilities.
In all, case 7 gives us
90 + 180 + 90 + 360 = 720 possible
cubes.
Do the same for each of the other cases
and use additivity to obtain the final result.
C.18 b) Hint: Consider disjoint cases.
C.19 a) Solution: There are 5 possibilities
for each digit. Therefore, by the
Multiplication Principle, there are
4^ = 625 four digit numbers.
b) Approach 1: Pair the numbers as
follows: Two numbers are paired if the sum
of the digits in a corresponding position is
io(e.g. 3757 is paired with 7353). The only
number which cannot be paired vdth
another number is 5555. Since from part
(a), there are 625 numbers in all, 624
numbers are paired, giving 312 pairs. The
sum of each pair of numbers is found to be
11110. Therefore, the sum of all the numbers
is 312(11110) + 5555 = 3471875.
Approach 2: Because of the symmetry
involved, each position of each digit appears
equally often. How often? Since there are
625 numbers and five distinct digits each of
which occurs in a position equally often,
each of the digits 1, 3, 5, 7, 9 will occur 625/5
= 125 times in each position (that is, there
are 125 numbers with a 1 in the last position:
125 with a 3, etc.) The units position adds to
125(1+3+5+7+9); the tens position sums to
10125(1+3+5+7+9); the hundreds position
sums to 100-125(1+3+5+7+9); the thousands
position sums to 1000125(1+3+5+7+9).
Altogether we get
(i+io+ioo+iooo)(i25)(25)=347i875.
c) All numbers with 1 in the first
Oeftmost) position precede all numbers with
3 in that position, which precede all
numbers with 5 in that position, etc. As
there are 125 numbers which have a 1 in this
position (see approach 2 above), 125 which
have a 3 in this position, etc. There are 250
numbers which start with a 1 or a 3, and 375
which start with 1, 3 or 5. The 314th number
must therefore start vnth a 5; in fact it is the
64th = (3i4-25o)th number in the list of
numbers starting with 5. We now look at the
second digit. Of the 125 numbers starting
vdth 5, 25 have 1 as the second digit; 25 have
3; etc.
Since 50< 64<75, the second digit of the
number we want must be 5. In fact the
desired number is the 14^^ = (64-5o)th
number in the list of numbers starting vnih
55.
Continuing in this manner we look at the
third digit of the numbers starting vdth 55.
There are 5 which have 1 as the third digit,
etc.
Since 10< 14 < 15, 5 is the third digit and
the fourth digit is the 4th = (i4-io)th
number with first three digits 555.
So the 314th number is 5557.
C.20. Hint: See the solution of C.19, but use
caution: The number of numbers with last
digit 1 is not the same as the number of
numbers vdth next to last digit 1, since 1
digit numbers are also included, etc.
Therefore, this problem should be broken
down into cases.
C.21. Hint: b) The digits are o, 1, 2, 3, 4, 5,
6, 7, 8, 9. However the first digit cannot be
o. Suppose we allow o to be a first digit
(e.g., as in 0132). How many numbers vnW
we then have? We can add up the sum of all
the numbers including these in a manner
similar to Exercise C.19. Then we subtract
the sum of the numbers starting vdth o.
C.23. Hint: Break up into cases depending
on the position in which the first red checker
appears.
C.24. Solution: a) In order for Delia to be
ahead when the fifth vote was counted,
either all four of her votes are among the
first five counted or three of her votes are
among the first five counted. In all there are
10
for the positions in which Delia's voted
could be counted. Of these there are
5^ fs^
m
which all four votes were
counted among the first five votes. ( The

424
HINTS AND SOLUTIONS
>i
indicates that none of her votes were in
Oj
the last five positions.) Similarly, there are
ways in which exactly three are
among the first two. (The
indicates that
'J
her remaining vote could be in any one of
the last five positions.)
Thus the probability that Delia was ahead
when the fifth vote was counted =
fs
4AoJ''l3Ai
10
= 11 /42 .
b) If she was ever ahead, then she was
ahead for the first time after the first, third,
fifth, seventh, or ninth vote was counted.
I.e., if she was ahead at the 2"^ vote, she
must have been ahead at the first, and if she
was ahead at the 4^^ vote, she must have also
been ahead at the third, etc.
The probability that she was ahead after
^9^
the first vote was counted is
4
10
The only way she could be ahead at the
third vote but not the first is if the first three
votes were VDD. There are
positions in
which Delia could receive her remaining two
votes, so that the probability that she is
ahead after the third vote but not after the
first is
1
10
There are two ways that Delia could be
ahead after the fifth vote, but not after the
first or third votes, namely if the votes went
VDVDD or WDDD. With either of these
(;]■
ways there are positions for Delia's 4^^
vote. So the probability that Delia was
ahead after the fifth vote but not sooner is
^5^
1
21
Similarly, the probability that Delia was
ahead after the 7*^ vote was counted, but not
sooner =
1
42'
Finally, the probability that Delia was
ahead after the 9^^ vote was counted = o,
since by that time Victor has at least 5 votes.
Since the above cases are disjoint the
probability that Delia was ever ahead is
4 111 4
—+ —+ — + — + 0 = -.
10 10 21 42 7
C.25. Hint: Consider cases: 1) five different
kinds of trees; 2) two trees of one kind,
other three are all different; 3) two pairs of
trees of same kind.
C.26. Hint: Consider all possible
placements of 2 kings on the board and
subtract the number of excluded or illegal
positions.
C.27. a) Hint: Consider cases depending
on size of the square. (I.e., count the
number of squares of each size)
b) Hint: A rectangle with horizontal
and vertical sides is determined by its upper
left hand and lower right hand corner.
c) Hint: Count the number of
squares of each size. Make sure that you
have included all possible sizes and
orientations. E.g., the two squares in Figure
49 are the same size but have different
orientations.
FIGURE 49

HINTS AND SOLUTIONS
425
C.28. Hint: See the hint for Exercise
C.27b).
C.29. a) Hint: What numbers can not be
made?
b) Hint: Label the dice (di, da, dg)
and hst the favorable outcomes.
C.30. c) Hint: Since we are only interested
in the positions of people relative to one
another, in a circular arrangement we may
hold one person fixed and consider the
seating of the other people in relation to this
person.
C.32. Hint: Consider the positions of the
yellow beads relative to each other. I.e., they
may be touching, there may be one, two or
three spaces between them. Count the
number of necklaces in each case. Be aware
of symmetry or lack thereof.
C.33. Hint: First consider the possible
distributions of colors, then the possible
arrangements for each particular color
distribution.
C.34. b) Solution: There are
ways of
selecting a 13-card hand. We now consider
cases as to whether there are exactly eight,
exactly nine, ..., exactly thirteen cards in a
particular suit.
We can select 8 cards in the given suit
(and 5 cards from the other suits) in
■ ways.
Similarly we can select exactly 9 cards
from the selected suit m
^^^^Jways.
Etc.
Since there are 4 ways of selecting the
suit with eight or more cards there are in all
39
5^
39
1
39
4^
'39^
0
favorable outcomes.
Therefore the probability that at least
eight cards in the same suit
d) Solution: Here we must be careful as
there may be more than one suit with six or
more cards. First, break into disjoint cases
where there are exactly six cards of a suit,
exactly seven cards, etc. The cases with
eight or more cards in a suit have been
handled in part b).
We can select seven cards in a particular
suit and six in the remaining three suits in
7 It) ^^y^-
We can select six cards in a particular suit
and seven in the remaining three suits in
' ways. As there are four ways of
choosing a particular suit it would seem that
the number of favorable cases is
This is not quite correct, as we have counted
some hands twice. For example, a hand
which contains 6 spades, 6 hearts, and 1
diamond is included when hands containing
6 spades are counted as well as when we
count hands containing 6 hearts. A similar
overcount occurs for a hand containing 6
spades and 7 hearts. Etc.
We therefore must subtract off the
number of hands that have been counted
twice, i.e., the number of hands that contain
six or more cards in two or more suits. This
happens in two types of hands - hands with
seven cards in one suit and six of another and
hands with six cards in each of two suits and
one card in a third suit.
In the 7, 6 case, there are four ways of
selecting the suit with seven cards and then
three ways of selecting the six card suit. The
seven cards can be chosen
'»['.']
ways; the
. . 13 . . 13 13
SIX m ways giving 4-3 ways of
getting a 7, 6 hand. In the 6, 6 case the two
six card suits can be chosen in ways
(since they are interchangeable); once the 3

426
HINTS AND SOLUTIONS
suits are determined, the cards can be chosen
'l3Yl3Y26^
" ways giving
ways of getting a 6, 6, i
6jy6)[ 1
4 V 13Yl3V26
1
2 A 6 A 6
hand.
Thus, the total number of favorable
outcomes is
13 ¥39
-4-3
131(13
To compute the probability this number
must be divided by
13
g) Solution: There are 4 ways to choose
the suit with four cards, ways to choose
these four cards,
13Yl3Yl3
ways to choose
. ^ A ^ A ^.
three cards from each of the other suits.
The probability of this case
A3Yl3Yl3Yl3^
C.35. c) Solution: There are possible
five card hands. A five card straight can
begin with one often cards. A, 2,..., 10. Each
of the five cards in a straight can be of any
suit, so the total number of favorable
outcomes is lO-4-4-4-4-4 = 10-4
,5
and the
probability of this event is
10-4"
(Note that
straight flushes are being included here.)
f) Solution: There are 13 possibilities for
the face value of the card repeated 3 times,
which leaves 12 possibilities for the face value
of the card repeated twice. The three suits of
the triply repeated card can be chosen in
4^
ways and the two suits of the doubly
repeated card can be chosen in ways,
giving the probability of this event =
'4Y4^
13-12
3 2
g) (As in e) there are 13 ways of
choosing the three of a kind. There are
12^
ways of choosing the face values of the
other two cards. The suit of each of these can
be selected in | | ways.
13
This gives the desired probability as
4Yl2Y4Y4^
3 Jl 2 / 1JU
52
j) Hint: Count the total number of
hands with cards of different face values,
subtract the number of straights and the
number of flushes but add back the number
of straight flushes because these have been
subtracted twice.
C.36. a) Hint: Since the five cards you have
been dealt are no longer in the deck, only 47
cards remain. If you keep only the two kings,
you must draw three cards so that there are
47^1
possible outcomes. You must also take
note that certain cards are no longer in the
deck.
For example, if you keep the two kings you
can obtain a full house in the following ways:
Case 1: Draw three Q's, three J's or three
lo's;
Case 2: Draw three cards of some other
face value;
Case 3: Draw one K and two Q's, two J's
or two lo's;
Case 4: Draw one K and two cards of
some other face value.
The difference between Cases 1 and 2 is
that only three Q's, J's, lo's remain in the
deck whereas four of each other face values
(Ace through 9) are in the deck.
Similarly for Cases 3 and 4.

HINTS AND SOLUTIONS
427
C.37. Solution: Call the first player B and
the second player C. B can beat you only by
holding 2 more A's. (B can't have a straight
since all the 5's are accounted for.) C can beat
you only if he has either two more 9's, two
more 8's, two more 6's, a 7 and a 10, or two
more hearts.
Consider the ways in which you will not be
the winner. Break these up into the following
disjoint cases:
Case 1: B beats you.
Case 2: You beat B, but C beats you
holding the AH as one of his hidden cards.
Case 3: You beat B, but C beats you not
holding the AH.
We compute the probability of each case.
In Case 1 there are | _ possible cards
that B could hold. Of these only | [are
favorable. So the probability of Case 1 =
.2) 6
37-36
In Case 2, C's fifth card must also be a
heart.
As there are only 6 possibilities for this
fifth heart, there are ways in which C can
beat you. In addition B can hold any two of
the remaining 35 cards except for both of the
^35
remammg aces
Thus there are
possibilities for B's hidden cards. Using the
Multiplication Principle, this case gives rise
to
favorable outcomes. And so
the probability of Case 2 =
vly
35
v2y
-1
37
v2y
24-594
37-36-35-34
In Case 3, C cannot hold any aces. There
are
^3^ f2^ f3Y3^ fe^
^3^
v2y
two 9's two 8's two 6's
= 31
possible hands that C could hold. For each of
these B can hold any two of the remaining 35
cards except for 2 A's. Again by the
Multiplication Principle, the number of
The probability of Case 3 =
4 3l[592]
favorable outcomes = 31
31
37 ¥35 I 37-36-35-34
The probability that you are not a winner is
24 • 594
4-31-592
37-36 37-36-35-34 37-36-35-34
7140+13536+73408
37-36-35-34
94084 94084
.059.
37-36-35-34 1585080
Therefore the probability that you win is
approximately I - .059 = .941.
C.38. a) Solution: Denote the people by P,
Q, R and the gifts they brought by p, q, r
respectively. We make a chart (Figure 50)
indicating the various receivers.
p
p
p
q
q
r
r
Q
q
r
r
P
P
q
R
r
q
p
r
q
p
unfavorable; P received p
unfavorable; P received p
favorable
unfavorable; R received r
favorable
unfavorable; Q received q
FIGURE 50

428
HINTS AND SOLUTIONS
Note that there are 3! = 6 possible ways of
distributing the gifts and that 2 of these are
favorable.
Note the number of favorable outcomes
could have been obtained as follows. It is
equal to 3! - the number of unfavorable
outcomes.
An outcome is unfavorable if exactly one
person receives his or her own gift or every
one receives his (her) own gift (it is not
possible for exactly two people to receive
their own gifts).
There is 1 way in which everyone receives
his or her own gift. Also if exactly one
person receives his or her own gift, the other
two won't. There are 3 ways to determine
which one person receives his (her) own gift.
Thus the number of favorable outcomes is
6-1-
1 = 2.
b) We could proceed by making a chart
as in a) but, as there are 24 cases, the
approach is cumbersome. We therefore
proceed as follows:
Let D(n) = the number of ways in which
n people can exchange gifts so that no one
receives his or her own gift. In this problem
we are looking for D(4). Note that in part a)
we found D(3) = 2 and it is clear that D(2) =
1.
To find D(4), note there are 4! ways of
distributing the gifts among 4 people. So
that
D(4) = 4! - number of unfavorable ways
of distributing the gifts. A distribution of 4
gifts is unfavorable if exactly one person
receives his or her own gift or if exactly two
people receive their own gifts or if everyone
receives his or her own gifts.
Exactly one person can receive his or her
own gift in D(3) ways (that is,
possibilities as to which person receives his
or her own gift, and, for each of these, there
are D(3) ways to distribute the remaining
gifts among the other 3 people so that no
one receives the gift that he or she brought).
Similarly, exactly 2 people can receive
their own gifts in D(2) ways. There is 1
way in which every one can receive his or her
own gift. Thus
D(4) = 4!-[jD(3)-r]^(2)-i
= 24-4-2-6-1 - 1 = 9-
c) As in b) D(5) =
5!-
D(4)-
%(3)r D(2)-i = 44.
d) Analogously D(6) = 265
By taking another approach D(n) can be
seen in general to be
D(n)= n!
,111 1
1 + + ...± —
2! 3! 4! n'
This result is obtained in the following
manner: The total number of outcomes
minus the number of outcomes for which
at least one person receives his or her
own gift plus the number of outcomes for
which at least two people receive their
own gifts minus the number of outcomes
for which at least three people receive
their own gifts plus, etc. (Think about
why we have to alternately add and
subtract these terms.)
e) We want the probability that Ross
and Sanford receive each other's gift
given that no one receives the gift that he
or she bought. If Ross and Sanford
exchange gifts then the other four can be
distributed in D(4) ways. Therefore the
desired probability is equal to
D(4) 9
D(6) 265
.033.
C.39. Hint: What is the connection with
Exercise C.38?
C.43. Solution: If the conditions of part
b) of Exercise C.14 are realized, then Jack
Smith is either sitting in the same row as
Joan Brown and John Jones or else he is
sitting in the same row as Joan Jones and
John Brown. Both cases are equally
likely to happen, so the probability that
Jack Smith and Joan Brown are sitting in
the same row is —. Given that they are in
2
the same row, there are 6 possible
arrangements only two of which are
favorable. Hence the desired probability
is
Pr[they are in the same row] • Pr[she is to
his immediate right/ Pr[they are in the
same row] =
1.1
2 3

HINTS AND SOLUTIONS
429
C.44. Solution: a) Since there are 49 cards
unaccounted for, 23 of which are six or less,
the desired probability is 23/49.
b) In this case 14 cards are still
unaccounted for, 4 of which are favorable.
Hence, the probability that if you draw a
card the total will not exceed 21 is — = - .
14 7
c) If you stand pat at 15, then, since the
dealer has a 9 showing, he will beat you if his
hole card is 8 or more. This occurs in 7 of
the 14 cases, so the probability that he will
already beat you is j . If his hole card is 7
or less, he will have to draw another card. If
he has a 7, (probability j^), then the
probability that he will not exceed 21 if he
draws is -^. Thus the probability of this
case IS
13 •
3 2
_3_
91
If he has a 6 (probability -j^), then the
probability that he will not exceed 21 if he
draws is j^. Thus the probability of this
case is -^ ~
_3_
91
The remaining cases - he has a 5 or he
has a 3 - give probabilities of tV "if ~ ^
and iV "n ~ ^ respectively.
Thus, the probability that the dealer will
beat you if you stand pat is
2 91 ^ 91 ^ 91 ^ 91 182 '
d) If you do draw, then the probabilities
for the card(s) the dealer holds and draws
will be affected by the card you draw. We
therefore consider cases as to what card you
might draw. (You should only draw one
card since any two of the cards left would
bring your total past 21.) If you draw a card
larger than a six (probability —) you
14
automatically lose.
If you draw a three (probability —),
14
your total will be eighteen.
There vnW be 13 cards still
unaccounted for. Of these, the dealer will
win if his hidden card is a ten or a picture
card or if his hidden card is a seven and
he draws a five, or his hidden card is a six
and he draws a five or a six, or if hid
hidden card is a five and he draws a six or
a seven. Therefore if you draw a three,
the dealer will win with probability
1-5 72
5 31
— +
13 1312
2-2
1312 13-12 156
84
156
Therefore
Pr[you vnn if you draw a three] ■
Pr[you draw a three and win]
-JL il-_?i_
"14 156 "2184
PrCyou draw a three and lose]
_ J 8 72
" 14'156 "2184
Similarly, you could draw a five with
probability —. In this case the dealer will
14
win with probability = —^ ; the
2_
156
dealer will tie
2-1 2
= ; and you
1312 156
13-12
you vdth probability
will vdn vdth
probability 1 = .
2 _152
156 156 ~ 156*
Therefore,
Pr[you draw a five and win]
1 152 152
14 156 2184
Pr[you draw a five and lose]
_ 1 2 _ 2
14 156 2184"
Pr[you draw a five and tie]
_ 1 2 _ 2
14 156 2184'
In a similar manner
1 . -. 300
PrLyou draw a six and v^n] =
2184
Pr[you draw a six and lose] = o
12
Pr[you draw a six and tie] =
2184
card the
Therefore if you draw a
probability that you win is
84 152 300 536
= + + = « .245;
2184 2184 2184 2184
the probability you lose is
10 72 2 1634
= — + + = « .748;
14 2184 2184 2184
and the probability that you tie is

430
HINTS AND SOLUTIONS
12
14
.006.
2184 2184 2184
Since the probability that you win if you
draw (.245) is less than the probability that
you win if you stand pat (.357), and the
probability that you lose if you draw (.748) is
greater than the probability that you lose if
you stand pat (.643), you should stand pat.
C.45. Hint: See solution to Exercise C.45.
case occurs
C.47. Solution: a) Since each of the 7
remaining people can put a "one" or a "two,"
there are 2^ possible outcomes. Of these
7^ (1
result in 4 ones and 3 twos, and
A) ^ -^ ' ^4
result in 4 twos and 3 ones. This gives
^7^
4J 35
.55.
2' 64
b) Given that the teams are selected,
there are possibilities for Dan's
teammates. Therefore the desired
probability is
c) There are four possible scenarios
which should be considered:
i) Dan and Eliot are both on the
same team as the designated player;
ii) Dan and Eliot are both on the
opposite team as the designated player;
iii) Dan is on the same team as the
designated player but Eliot is not;
iv) Eliot is on the same team as the
designated player but Dan is not.
In each case, there are five players
remaining.
In case i), one of the five must be on the
team vdth Dan and Eliot, so this case can
occur in =5 ways; in case ii), two of the
five must be on the team v^th Dan and Eliot,
so this case can occur in = 10 ways; in
case iii), two of the five must be on the team
with Dan and the designated player, so this
10 ways; and in case iv).
two of the five must be of the team with Eliot
and the designated player, so this case
occurs in =10 ways. Since cases i) and
ii) are favorable, the required probability is
5 + 10 15 3
5 + 10 + 10 + 10 35
C.49. Hint: What is the probability that no
two of them have the same birthday?
C.54. Hint: Play the movie in reverse.
C.62. Solution: Clearly, 1 must be placed in
the upper left-hand corner and 8 must be
placed in the lower right. Figure 51 shows
all possible placements of 2, 3 and 4.
1
2
3
4
8
(a)
1
3
2
4
8
(c)
1
3
2
4
8
1
4
2
3
8
(b)
1
2
3
4
8
(d)
1
2
3
4
8
(e)
FIGURE 51
(f)
In (a), the placement of 5, 6, and 7 is
forced, (a) gives rise to 1 case (b), (c) and (d)
each give rise to three possibilities in each
either the 5, 6 or 7 can be on the top row. In
(e) and (f) the placement of 5 is forced;
therefore each gives rise to two possibilities
depending on the placement of the 6 and 7.
1 + 3 + 3 + 3 + 2 + 2 = 14
(a) (b) (c) (d) (e) (f)
C.63. Hint: Consider the possible patterns
for the placement of 1, 2, 3 and 4. Separately
consider the possible placements of 6, 7, 8
and 9. How can these patterns be
numbered?
C.64. Hint: Label each hot dog by the
number indicating the order in which it will

HINTS AND SOLUTIONS
431
be eaten. That is, the first hot dog to be
eaten will receive a i; the second, a 2, etc.
In how many ways can these numbers be
assigned to the hot dogs in the leftmost
column? Once this column has been
assigned numbers, how many ways can
numbers be assigned to the middle column,
etc?
C.66. Solution: There is only 1 way to go
from A to B; there are only two ways to go
from A to C (A^C or A^B^C); there is
only one way to go from A to D.
Therefore there are 1 + 2 + 1 = 4 ways to
go from A to E since we must get to E
directly from either B, C or D.
Continuing in the same manner we can
label each vertex of the diagram with the
number of ways in which that vertex can be
reached from A. The label of each vertex is
the sum of the labels of the vertices which
could immediately precede the vertex in
question.
C.67. Hint: See the solution of Exercise
C.66.
C.68. Hint: Label each letter by the number
of ways it can be reached. See the solution
of Exercise C.66.
C.69. Hint: First, count the number of
routes to I in a manner similar to that of
Exercise C.68.
C.70. Hint: See the solution of Exercise
C.66 and the hint to Exercise C.68.

Answers to
Selected Problems
CHAPTER 1
Exercises
1.1. Sue*s mother (Jo) owns the garter snake.
1.4. Allen is the third baseman; Bill is the center
fielder; Chuck is the catcher; Ed is the shortstop;
Harry is the pitcher; Jerry is the second baseman;
Mike is the right fielder; Paul is the first baseman;
Sam is the left fielder.
1.6. Tuesday.
1.7. Freddy came in last in the high jump.
1.8. Alice beat Bob, 9-1; Alice beat Carol, 7-3;
Alice beat Ted, 10-0; Bob beat Ted, 6-4; Bob tied
Carol, 5-5; and Carol beat Ted, 8-2.
1.10. Partial answer: Patti beat Nancy, 5-0; Patti
beat Manny, 5-0; Patti beat Tom, 5-2.
1.11. Cleo was the big winner. They are seated in
alphabetical order clockwise.
1.12. Leopold and Eli.
1.13. Mr. Lawyer—the plumber.
1.16. John ordered three bologna and received one
salami and two bologna.
1.17. They are labeled $.40 and $.45; they contain
$.75 between them.
1.18. Bobbie is the cheerleader; Frankie, the class
president; Gerry, the principars child; Jo Jo, the
basketball player; Ronnie, the volleyball player; Sal,
the valedictorian.
1.19. (a) Helen and Max play for East; Sylvia and
Lee for West; Emily and Ted for North; Vicki and
Paul for South; Becky and Irv for Central.
(b) Thursday: South beat Central, North beat East,
and South beat North; Friday: South beat Central in
the first match; then Central beat South to win
the championship.
1.20.
Competitor
Jimenez
Harris
Boone
Manners
Twofeathers
School
Mamaraneck
Jupiter
West Roch.
Marmaduke
Holbrook
Entry
number
5
2
1
3
4
Prior
standing
4
1
3
2
5
High
jump
4
5
2
3
1
1.21. The Norwegian drinks water; the Japanese
owns the zebra.
433

434
ANSWERS TO SELECTED PROBLEMS
1.22. "Kumquats" was at The Sarcophagus;
" Labradorian" at The Thumbscrew; "Mahatma"
at The Reptilian; "Nonconformist" at The
Purgatory; and "Ottoman" at The Quagmire.
Who went with whom to what show can then be
determined from the chart below:
Albert
Barney
Chuck
Danny
Ernie
Florence
O
Thur
K
Sat
mat
N
Fri
L
Wed
M
Sat
nite
Glenda
N
Sat
nite
M
Thur
K
Wed
O
Sat
mat
L
Fri
Helen
M
Wed
O
Fri
L
Thur
K
Sat
nite
N
Sat
mat
Inez
K
Fri
L
Sat
nite
M
Sat
mat
N
Thur
O
Wed
Joan
L
Sat
mat
N
Wed
O
Sat
nite
M
Fri
K
Thur
For example, the upper left box reads: Albert and
Florence went to see the "Omnipotent Ottoman"
on Thursday.
CHAPTER 2
Practice Problems
2.A 1. (a), (d), and (e) are statements.
3. (a) This sentence is true for some values of x
(x = 3) and false for others. But, as x is not a
specific number, we cannot say that the sentence
is true or that it is false.
(b) If it were a statement, it would have a truth
value. But both possible truth values lead to
contradictions. That is, if the sentence is true,
then it must be false (since it claims to be false),
but if it is false, then it is a true statement.
2.B 1. (a) Since both the antecedent and the
consequent are true, the statement is true.
(b) Since the antecedent (Jarvis got an F on the
final) is false, the statement is true.
(c) Since the antecedent (Jarvis got an F on the
final) is false, the statement is true (regardless
of whether or not Jarvis passed the course).
(d) Since the antecedent (Jarvis passed the course)
is true and the consequent (Jarvis got an A on
the final) is false, the statement is false.
(e) Since the antecedent (Jarvis failed the course)
is false, the statement is true.
3. (a) / -► f. If you recently fired a gun, then the
paraffin test is positive.
(b) (~r)—►^. If it doesn't rain tomorrow, then I
will go to the beach. Or, equivalently (^g) -*■ r;
If I don't go to the beach tomorrow then it will
be raining.
(c) h^ b. If I hear you say so, then I believe it.
(d) a-> /. If you ask me nicely, then I'll leave.
(e) s^c. If Tex sees a cowboy movie, then he
cries.
5. (a) p -^ q. This is true if Alice and Bob both
passed math, if they both failed math, or if Alice
failed but Bob passed.
{b)q-^p. This is true if Alice and Bob both
passed math, if they both failed, or if Bob failed
but Alice passed.
(c) p /\q. This is true if Alice passed math and
Bob also passed math.
7. (a) Jack is a good golfer if and only if he is not
a good tennis player.
(b) Jack is a good golfer, but he is not a good
tennis player.
(c) Jack is a good tennis player.
(d) Jack is not a good golfer.
(e) If Jack is not a good tennis player, then he
is a good golfer.
2.C 1. (a) We can conclude that no one was home.
(b) We can conclude that Archie was not early.
2.D 1. (a)^-(~/)). (b) 5-[(^/))vr].
(c) r-p. (d)(r^/))-[5A(^^)].
3. (a) If the poet is dreaming, then Titania is not
the queen of the fairies.
(b) Either the poet is not dreaming, or else Titania
is not the queen of the fairies.
(c) It is not possible that both "the poet is
dreaming" and "Titania is queen of the fairies"
are true.
(d) If it is not true that the poet is either
dreaming or jesting, then if Oberon is the king of
the fairies then Titania is the queen of the fairies.
5. (a) No one is home.

ANSWERS TO SELECTED PROBLEMS
435
(b) Archie is not early and Janice is not absent.
(c) No one is home.
7. (a) Either three krimmls are not worth one glunk
or the Martian canals are not empty.
(b> Three krimmls are not worth one glunk.
(c) I am not a Martian.
(d) Either the Martian canals are not empty or I
am not a Martian.
(e) We cannot conclude anything else. (The
statement is true regardless of whether Xcag Zemph is
the ruler of Mars or not and regardless of whether
or not the canals are empty.)
/vc
/
(d) Premises:
P,: If Lee Wong is a student, then he is bright.
P2: If Lee Wong is bright, then he studies logic.
Conclusion: If Lee Wong is a student, then he
studies logic.
^l
2.E 1. (a) If Jonas does not have at least ten
cents, then he does not have two nickels in his
wallet.
(b) If someone could enter, then the door is not
locked.
(c) If I won't hear some music, then I won't go
to the concert.
(d) If a triangle is not equilateral, then its three
angles are not equal.
3. (b) and (c) are equivalent; (a) and (e) are
equivalent.
2.F 1. (a) Premises:
P^\ Either Rachelle is brilliant or she has a
wonderful personality.
P2: If Rachelle has a wonderful personality, then
she has an active social life.
Conclusion: Either Rachelle is brilliant or she has
an active social life.
bwp
.'. bva
(b) Premises:
Pi: If Bessie is a cow, then she moos.
P2: Bessie is not a cow.
Conclusion: Bessie does not moo.
c -^ m
(c) Premises:
P^: Either Lucretia is forceful or she is creative.
P2: Lucretia is forceful.
Conclusion: Lucretia is not creative.
. (a) Valid. (b) Valid.
(c) Invalid (/> and q may both be true).
(d) Invalid (p and q may both be false).
. (a) Valid.
(b) Invalid. Bessie might be a bull, or even a
person who moos. In this case, both premises
would be true but the conclusion false.
(c) Invalid. If Lucretia were forceful and creative,
then both premises would be true but the
conclusion would be false.
(d) Valid.
. (a) Invalid, (b) Valid.
Exercises
2.2. The first was truthful and the second was a
liar.
2.4. The fourth troll is a Truthful and the others are
Liars.
2.5. The tall troll is Lowax; the short one, Waldar;
and the medium one, Gaut.
2.7. Winken is a Liar; Blinken is a Truthful; and
Finken is an Alternator who is lying.
2.10. The first robot is a Lawbake and the second
is a Mendible. Captain Cooke was using telic coins.
2.13. Sykes did it. Tolliver was also present.
2.15. Fifi.
2.17. He is under 5 feet tall, has white fur, and
5 toes on each foot.
2.18. Dr. Mandlebaum is the hematologist; Coates,

436
ANSWERS TO SELECTED PROBLEMS
the endocrinologist; Mavis, the cardiologist; and
Rowe, the gastroenterologist.
2.19. Danny received his present from Jason;
Ethan received his from Danny; Mark received his
from Ethan; Mindy received hers from Mark; and
Jason received his from Mindy.
2.20. Earl is the mason, Moe is the carpenter,
Luis is the plumber, and Randy is the architect.
2.22. Amy bought comic books; Betty bought a
coloring book; Carmen bought the ball; and Dee
bought candy.
2.24. Martians do not have three heads, they are
green, they can fly.
2.25. His mechanic. He also should replace his
spark plugs and distributor cap.
2.27. A blue shirt and white socks (and a brown
suit—if he is wearing any suit at all).
2.29. No, Luigi's plane could not fly more than 25
feet high.
2.30. Yes, Pamela's porridge is putrid.
2.32. Attila did not die at the age of seventy-nine;
the emotional development of primates does not
parallel that of reptiles; we cannot tell whether or
not Attila was reincarnated as a snake.
CHAPTER 3
Practice Problems
3.A 1. (a) «; « + 15. (b) n; 2n - 3.
(c) «;«(«-2). (d)d;\d. (e)a;-.
a
3.B 1. (a)« = «2-3. (b)«2 = „+i2.
(c) « + 1 < \n\ (d) / > 5 + w.
Exercises
3.2. $91.90.
3.4. 54 pieces.
3.5. Tom should get $6.30 and Don $2.10.
3.8. 200 yards x 900 yards, and 450 yards x 400
yards.
3.10.
1 + x/5
3.11. x=5.
3.12. $.50.
3.13. $109.40.
3.14. $5.40.
3.15. Ada weighs 299; Brendan, 348; Corinne, 327;
Darryl, 272; Eva, 311; and Floyd, 370.
3.17. Minski spent $17 and his wife spent $34;
Pinski spent $11 and his wife spent $40; Dubinski
spent $36 and his wife spent $15.
3.18. The 50 honor note.
3.19. 78.
3.21. 64.
3.22. 76 people (the baby wasn't a person when it
left Fort McConnell).
3.25. Chin Lee is 28, Sue Ling is 42.
3.26. Mutt is 40, Jeff" is 50.
3.28. The butcher is 48, the baker is 42, and the
candlestick maker is 60.
3.31. The hare can't win. Just as he is finishing
the first half of the race, the tortoise is crossing
the finish line.
3.32. 3 seconds.
3.33. 10:55 am.
3.34. 168 miles.
3.36. 16 nautical miles.
3.37. 55 miles.
3.38. (a) 350 yards, (b) The cyclist who starts at
the southern end. (c) At the southern tip of the
track.
3.39. 40 miles per hour.
3.41. 2 miles per hour.
3.42. 60.

ANSWERS TO SELECTED PROBLEMS
437
3.43. 45.
3.44. Yes (see Hints and Solutions section).
3.45. 18 stations; 16 minutes.
3.46. 3| miles per hour.
3.48. (a) 25^ minutes.
3.49. 21 minutes.
3.50. 32i
3.5 L They save 1 hour 22J minutes.
3.52. (a) 50 days, (c) Forever.
3.53. I am a male doctor.
3.55. Neither. The amount of oil in the vinegar was
equal to the amount of vinegar in the oil.
3.56. (a) 3^ gallons.
3.58. Beulah (260 lb) is married to Bobby (260);
Barbra (160) is married to Whelan (320); Belinda
(200) is married to Wally (300). Bubba is single and
weighs 280.
3.59. 5.
3.61. (a) The ratio of the length of the right arm
to the left is 6 to 5. (b) 20 oz. (c) 10 oz.
3.62. ^.
3.66. 36.
3.68. There are 28 beggars and I have 220 cents.
3.69. 7.
3.71. 8 and 5.
3.72. He turned the entire packet over behind his
back.
3.73. 31.
3.74. When the card was replaced, he counted off
7 cards to make sure that the selected card was the
eighth card of his half.
3.75. He peeked at the original bottom card of the
deck and predicted that card.
3.77. He counted to the 32 - 9 = 23rd card.
3.79. The four of diamonds was the top card of
the deck at the beginning of the trick.
CHAPTER 4
Practice Problems
4.A 1. (a) Since k = 6(x + 2y), then by part (ii)
of Theorem 4.1, 2\k, 3\k, and 6\k.
3. a + b = c. Suppose d\a and d\b. Then a = de
and b = df, for some integers e and /.
c = a + b = de + df = d(e + f).
Since g +/ is an integer, d\c.
Similarly, if d\a and d\c, then a = de and c = df
for some integers e and /.
b = c - a = df - de = d(f - e).
Hence, d\b.
The remaining case is similar.
4.C 1. (a) 15120 = 2^ -33 -5 -7.
(b) 2183 = 37 • 59.
(c) 409 = 409.
(d) 72814 = 2 • 7^ • 743.
4.D 1. (a) 15120 = 2^ • 33 • 5 • 7 is neither a
perfect square nor a perfect cube.
(b) 46656 = 2^-3^ is both a perfect square and a
perfect cube.
(c) 1728 = 2^ • 3Ms a perfea cube but is not a
perfect square.
3. (a) ^46656 = 2^ • 3^ = 8 • 27 = 216.
(b) ^^46656 = 2^ • 3^ = 36.
yi728 = 2^ • 3 = 12.
5. (a) X must equal 11 times a perfect square.
(b) y must be a perfect square.
7. A number m = p['p2^ "Pa* is a perfea nth
power if each exponent of each prime is divisible
by fi—that is, h |r, for all i = 1,2,...,/?.
4.E 1. Fifty.
3. (a) 1, 2, 4, 7, 8, 14, 28, 49, 56, 98, 196, 392.
(b) 1 and 353.
(c) 1, 2, 4, 7,8, 14, 16, 28, 56, 112.
5. (a) 105. (b) It is odd. (c) It is even.
4.F 1. No; gcd(3, 6) = 3 but 3 I 5.
3. k is divisible by 3.
4.G 1. (a)(i)4; (ii) 2; (iii) 1; (iv) 3.
(b) 37 and 54 are relatively prime.
3. (a) 3\
(b) p"*, where m is the smaller of r and s.

438
ANSWERS TO SELECTED PROBLEMS
(c) y • 5K
(d)/>r/>r"-/>r-
(e) 2^ • 32 • 5 • 7 = 2520.
4.H 1. quotient = 36, remainder = 0.
3. q= \2,r = 3.
5.
4.1
3.
4.J
3.
5.
7.
4.K
3.
Q =
1.
12;
1.
2.
20.
X =
. 1.
- 19, r
0={..
T={..
2={..
3={..
35; «.
(a), (b)
3)fe+ 1,
X + y
xy
= 2.
., -8,
., -7,
., -6,
., -5,
-4,0,4,8,
-3, 1,5,9,
-2, 2, 6, 10
-1, 3, 7, 11
12, ..
13,..
, 14,.
, 15, .
(d), and (f) are true.
k = 0.
= \\ =
= 30 =
(a) 4x = 4 (mod 5)
(b) 2x = 2 (mod 3).
5.
(a)y = \ (mod 7).
(b) -
7.
9.
4.L
1.
X =
1.
4.M 1
3.
5.
2x + 2y
2.
(a) 5.
. 8.
±1, ±2, ...
3 (mod 8).
6 (mod 8).
or -X = —
= 2 (mod 7).
(b) 1.
(c) 3.
.}
.}
..}
..}.
[ (mod 5)
4.N 1. (a) X =5 (mod 81).
(b) X = 5 (mod 12).
4.0 1. (a) One (mod 8).
(b) Three (mod 9), but only one (mod 3).
(c) None.
(d) Two (mod 98), but only one (mod 49).
3. (a) X = 11 (mod 12).
(b) There aren't any solutions.
(c) X = 2, 6, 10 or 14 (mod 16); or, equivalently,
X = 2 (mod 4).
(d) X = 1 (mod 3).
4.P 1. (a) X =3 (mod 12).
(b) X = 11 (mod 12).
(c) X = 38 (mod 59).
4.Q 1. X = 66,>; = 59; and x = 85, >; = 76.
3. X = - 4 + 53k, y = -3 + 31k, for k = 0, ±1,
±2,....
Exercises
4.2. 118.
4.4. $10,737,418.24.
4.5. 648.
4.6. 12.
4.7. There are 5 boys and 2 girls; the parents
have been married for 11 years.
4.9. Quincy owned 15 chickens; his sister, Trixie,
owned 5. Ralph owned 14 chickens; his sister. Vera,
owned 7. Pedro owned 13 chickens; his sister, Sandy,
owned 12.
4.10. a = 144, b = 1728, c = 169, d = 2197.
4.12. The ages are 6, 6, and 1. (Note the two
oldest are twins.)
4.15. $34.41.
4.18. Two.
4.19. $1.22.
4.20. She sold 5 kitchen clocks, 3 cuckoo clocks,
and 2 grandfather clocks.
4.24. The magician would have said, "Count up
to TV'' (where TV is 12 more than the second
number).
4.26. By placing the proper card in the tenth
position before the trick began. (Use algebra and
digital roots.)
4.29. 2.
4.30. 61; Sean got the best deal. The owner of
the gum ball machine got the worst.
4.31. $4.50.
4.33. (a)
Divisor
2
3
4
5
6
7
Smallest
123456798
123456789
123457896
123467895
123456798
123456879
Largest
987654312
987654321
987654312
987643215
987654312
987654213

ANSWERS TO SELECTED PROBLEMS
439
Divisor
8
9
10
11
12
13
14
15
16
Smallest
123457896
123456789
There are none
123475869
123457896
123456879
123457698
123467895
123457968
Largest
987654312
987654321
There are none
987652413
987654312
987654213
987653142
987643215
987654312
4.35. Six.
4.36. (c) Tuesday. (d) 1920, 1948, 1976.
(e) None, (f) Sunday, Wednesday, and Friday.
(g) 1905, 1911, 1916, 1922, 1928, etc. (mod 28).
(h) 171.
CHAPTER 5
•
0
1
2
3
4
5
6
7
0
0
0
0
0
0
0
0
0
1
0
1
2
3
4
5
6
7
2
0
2
4
6
10
12
14
16
3
0
3
6
11
14
17
22
25
4
0
4
10
14
20
24
30
34
5
0
5
12
17
24
31
36
43
6
0
6
14
22
30
36
44
52
7
0
7
16
25
34
43
52
61
3. (a) (11211Xhree.
5. (a) (lllOlllOXwo.
7. (a) (1112112Xhree.
(b)(lll2Xhree.
(b)(100011Xwo.
(b)(111101Xhree.
5.D 1. (a) C= 1.
(b) F is larger than B. (AB < 100, so F • AB <
F • 100.)
(c) B = 3 and F = 5 or B = 6 and F = 8.
(d) If B = 3 and F = 5, then A = 6 or 7; if
B = 6 and F = 8, then A = 7.
(e) B = 3, F = 5, A = 7. 1095 ^ 73 = 15.
Practice Problems
5.A 1. (a) 13. (b) 344.
(c) 3.833... = 3|. (d) 1299.
5.B 1. (a) (100000100lll)two.
(b)(2212022)three.
(c) (13355)3ix.
(d)(6041)sevcn.
(e) (125E),welve.
3. (a) (lO)two. (b)(10)three. (c) (10)b.
5.C 1.
+
0
1
2
3
4
5
6
7
0
0
1
2
3
4
5
6
7
1
1
2
3
4
5
6
7
10
2
2
3
4
5
6
7
10
11
3
3
4
5
6
7
10
11
12
4
4
5
6
7
10
11
12
13
5
5
6
7
10
11
1?
13
14
6
6
7
10
11
12
13
14
15
7
7
10
11
12
13
14
15
16
Exercises
5.1. He btcame numb Burr, Count Turr, after
Wiiting the 9 in 8196.
5.2. (b) Any digit could be the last digit of a cube.
5.4. R. Velt.
5.6. 11, 12, 15, 24, and 36.
5.9. (b) 11, 12, 15, 22, 23, 26, 33, 34, 37, 44, 45,
48, 55, 56, 59, 66, 67, 77, 78, 88, 89, 99.
(c) 11, 14, 22, 25, 33, 36, 44, 47, 55, 58, 66, 69,
77, 88, 99.
5.11. The only such number is 65.
5.13. He wrote 1089 on the index card before the
trick began.
5.14. December 4.
5.15. 10.
5.16. 255. Tweedledee should always guess the
number in the middle of the remaining permissible
range.

440
ANSWERS TO SELECTED PROBLEMS
5.18. He added the numbers in the upper left hand
corners of the indicated cards.
5.19. (b) It reverses the order.
(c) After four deals the cards would be back in
their original order.
5.23. (a) 22.
5.25. Between 1 and 13, weights of 1, 3, and 9
grams suffice; between 1 and 40, weights of 1, 3, 9,
and 27 will do; in general, one weight for each
power of 3 up to the desired total will be needed.
5.27. 2.
5.29. 19.
5.30. If the base is even, n is odd if and only if
the last digit (the units digit) of its representation
is odd; if the base is odd, n is odd if and only if
the sum of the digits in the representation of n is odd.
5.33. MY.
5.34. MAID = 6423.
5.35. A nag.
5.37. 14579 + 85919 = 100498.
5.39. 125 • 37 = 4625.
5.40. 246 • 386 = 94956.
5.45. 37 • 29 = 1073.
5.46. 74 • 59 = 4366.
5.47. CHEIN = 38921; AVERBACH =
76905738.
5.50. 12800874 ^ 142 = 90147.
5.52. 101010101 ^ 271 = 372731.
5.53. 3040774 ^ 178 = 17083.
5.55. The two divisors are 333 and 29 respectively;
the two quotients are 300324 and 10356 respectively.
5.56. 631938 h- 625 = 1011.1008.
5.58. 152152 is the only one.
5.60. Any base greater than or equal to five.
5.61. Base seven or base eight.
5.63. Base four. 102003 ^ 33 = 1031.
CHAPTER 6
Practice Problems
6. A 1. (a)r
(b)
vertex
degree
vertex
degree
A
2
B
4
C
3
D
3
E
4
1 ^
1 ^
B
5
1 ^
1 ^
D
3
f'
4
6.B 1. (a) ABCDEFGHIBJCEJFHJIA
(b) ABCFEDGHIFHEBDA
(c) None exists.
6.C 1. (a) BACFJIHGDBCEFIEHDEB
(b) FIHGDABCFEDBEH
(c) None exists.
6.E 1. (a) ABGDFADB; CHGEBCEH;
MLGJHMJL; and KFGILKIF. Four are needed.
(b) ABHGACDFEC; BD; FH; and GE. Four are
needed.
3. (a) ABGDFADBCHGEBCEHLKGIHLIKJ-
FGHKJHF. Three retracings are necessary.
(b) ABJGACDBDFJFEGEC. Three retracings
are necessary.
5. (a) ABGDFADBCHGEBCEHLKGIHLIKJ-
FGHKJHFA.
(b) ABJGACDBDFJFEGECA.
6.F 2. (a) Not possible. There are too many odd
vertices.
(b) Dicircuit: BCFIHGDABFEDHEB.
(c) Not possible. E has in-degree 4 and out-
degree 2.
6.G 1. (a) Four colors.
(b) Four colors.
(c) Two colors.
(d) Three colors.
3. As this figure shows, the number of colors
needed are (a) three, (b) five, (c) four.
(a)
(b)

ANSWERS TO SELECTED PROBLEMS
441
n:n
n t"i
i 1 i i
color 1
color 2
color 3
- - color 4
color 5
(c)
Exercises
6.2. Colorado.
6.5. (a) No. (b) « = 3.
(c) AGIBHJCLFDKEKDFLCJHBIGA.
6.7. ACFDCFGEDGEBDAB; 1400 meters.
6.8. (a) 3350 meters. Her route could be:
DABDEBCFBFEHDGHFIH.
6.11. 111211312322122233213331323.
6.13. There are two different routes plus their
reversals.
6.16. THISISTOODIFFICULT or
THISISODIFFICULTOT.
6.17. No.
6.19. (a) Yes, it is possible.
(b) It is not possible.
(c) It can be done on an « x « board if n is even
but not if n is odd. More generally, it can be done
on an w X « board unless m and n are both odd, in
which case it cannot be done.
6.28. (b)
Trip
number
1
2
3
4
5
6
7
Resulting state
Near shore
hjhjhjh^WiWj
hihjhjh^WjWjWj
(kayak)
hihj hj h^ w,
hjhj hj h^ WjW,
(kayak)
hjhj WjWj
(kayak)
hjh^w,
Island
W3W4
(kayak)
W4
W4
W4
W4
W4
W4
Far shore
Wj W3(kayak)
W3
h, h^ W3 (kayak)
hsW,
(kayak)
8
9
10
11
12
13
14
15
16
h.h^w,
hjh^WiW^
(kayak)
w,w.
W,W4
WjW^
WjWj W4 (kayak)
w,
WjWj(kayak)
W3W,
(kayak)
W3
W3
W2W3
(kayak)
h^hjW^
h^hjW^
(kayak)
hjhjhjh^
hjh^hjh^WjWj
(kayak)
hjhjhjh^Wj
hjhjhjh^WjWjWj
(kayak)
hjhjhjh^WjW^
everyone
6.29. 20.
6.30. Hugo descends and gets out. The chest is
lowered and Hugo gets in with the chest. Jon
descends as Hugo and the chest rise. Hugo and the
chest are removed at the top, and Val descends—
causing Jon to rise. The first three steps are
repeated and Jon gets out. At this point Jon and
Val are at the bottom and John is in the tower with
Hugo, the chest, and the key. Now Hugo descends
and gets out, then the chest is lowered. John descends
while Val, Hugo, and the chest ascend. (If the basket
will hold only two people at a time, Jon and Val
ascend while John and Hugo descend, and the rest
of this answer is modified slightly.) Val removes the
chest and Hugo, and descends while Jon ascends.
The first three steps are then repeated again. Hugo
gets out at the top and Jon gets out at the bottom,
causing the chest to come tumbling down. Hugo
picks up the key in his mouth and drops it out of
the window. He then descends.
6.35. It can be done in sixteen moves.
6.36. Twenty-two moves.
6.37.
t • • *
1 D
T • • *
T—•—•—'
T t • '
•
i
1—•—i
1 B
1 • • *
ric
1 (
)—•—i (
• 1 B»
» i • <
» •
•
•
•
»—•
*D
fE

442
ANSWERS TO SELECTED PROBLEMS
6.38. No.
6.40. Amos, Crawford, Everett, Burl, and then
Dirk.
6.43. There are two possible final positions (see the
figure below). They may be reached in several
different ways.
Q
Q
^
^
Q
\
x*
Q
Q
->
Q
Q
Q
(a)
Q
Q
Q
y'
V
Q
Q
^
•v
Q
s
^
Q
Q
(b)
6.46. 18.
CHAPTER 7
Practice Problems
7.A 1. (a) Chance is involved; the game is of
perfect information and is finite.
(b) The game is finite and decisionmaking is
chance-free, but it is not a game of perfect
information.
(c) Backgammon is a game of perfect information,
but it involves chance and is not finite.
(d) Gin rummy involves chance, is not a game of
perfect information, and is not finite.
(e) Chinese checkers is a chance-free game of
perfect information, but it is not finite.
(f) Tic Tac Toe is a finite, chance-free game of
perfect information.
(g) Parcheesi is a game of perfect information,
but it is not chance-free and is not finite.
7.B 1. Tic Tac Toe may end in a draw. The
argument does show that the first player does have
at least a drawing strategy.
2. The argument used for Hex shows that A has
at least a drawing strategy, since A's earlier moves
can't hurt him later on.
7.F 1. (a), (b), and (d) are equivalent.
3. Placing an X in the upper righthand comer.
7.G 1. The positions in (a) are equivalent; so are
the positions in (c). The positions in (b) are not
equivalent.
7.H 1. (a) Yes; fill in the middle column.
(b) Yes; fill in the top row.
(c) No.
(d) Yes; complete the second column.
7.1 1. (a) Losing. (b) Winning.
3. (a) Take 7 from the pile of 8, leaving (1, 6, 7).
(b) Leave (7, 11, 12), (5, 11, 14) or (2, 12, 14).
(c) Give up.
(d) Take 10 from the 11, leaving (1, 2, 4, 7).
Exercises
The answers below either present a list of the winning
positions in the game or give the identity (A or B)
of the player who wins. Occasionally, when A is the
winner, a correct first move is also indicated. (Other
first moves might also be correct.)
7.1. (b) (i) Winning positions: n = 6k; k = 0, 1,
2, ....
(b) (ii) Winning positions: n = 6k + 2; k = 0,
1,....
7.3. (i) Winning positions: n = (m + j)k\ /? = 0,
1, ....
(ii) Winning positions: n = (m + j)k -\- j; k = 0,
1, ....
7.4. (a) (ii) A wins by taking 4.

ANSWERS TO SELECTED PROBLEMS
443
(b) (ii) Winning positions: n = 7k + I and
7k + 3; k = 0,\, ....
7.7. (i) Winning positions: n = SkorSk + 1 if you
possess an odd number, and Sk -\- 4 or 8/j + 5 if
you possess an even number.
7.8. (i) Winning positions: 0", k)J <k where k -j
is divisible by 4.
(ii) Winning positions: (4j, 4k -h 1) or (4j + 1, 4k)
or (4; + 2, 4k-h 2) or (4; + 3, 4)fe + 3), ;, k = 0,
1,2,....
7.9. (a) (i) and (ii) A wins by taking 2 sticks from
the first pile to leave (9, 15).
7.11. B wins.
7.12. (a) (ii) A wins by leaving (3, 3, 3).
(b) (ii) Winning positions: (0, 0, 1) and (k, k, k)
for yfe = 2, 3, 4, ....
7.13. (a) A wins by leaving (2, 5, 7, 7), (3, 4, 7, 7),
or (3, 5, 6, 7).
(b) Winning positions: Those for which all
column sums are divisible by 3 when the numbers
of sticks in each pile are written in the binary system.
7.14. Winning positions: Those for which all
column sums are divisible by m + 1 when the
numbers of sticks in each pile are written in the
binary system.
7.15. (a) (i) A wins by wiping out any pile.
(b) (i) Winning positions: (0, even, even),
(0, odd, odd); all other positions are losing.
7.16. (a) A wins by taking the middle stick.
(b) A wins by taking one end stick.
(c) A wins by taking either the middle stick if n
is odd or the middle two sticks if n is even.
7.19. (a) The first player wins by making piles of
10 and 4 sticks.
(b) The first player wins by making piles of 12
and 3.
7.20. (a) Winning positions: leave two odd piles.
(b) Winning positions: leave one pile containing
1 or 2 sticks and the other pile containing ?>k + 1
for some /? = 0, 1, 2, ... .
7.21. A wins by taking an ace.
7.22. A wins by turning up a 4.
7.24. A wins by selecting 24, 25, 28, or 29 (or any
number of the form 9^+1, 9)fe + 2, 9/j + 6, or
9)fe + 7 for /j = 1, 2, ... .
7.25. (a) November 30. (b) January 20.
7.26. B wins by always staying on the same
diagonal as A.
7.27. A wins by moving four cells diagonally and
then using a pairing strategy indicated by the lines
in the figure below. That is, if B moves to a cell
with a line ending in it, A moves to the cell at the
other end of that line. A could also win by moving
two or three cells diagonally, but it would take
longer.
\
\
\
\
\
\
\
^L^A
nT:
^B
7.28. (a) A wins by moving any one of his
counters as far to the right as possible.
(b) If m is odd A wins; otherwise B wins.
7.29. (a) A wins by starting in the center.
(b) A draws by starting in the center and then
playing symmetrically opposite to B, using the
symbol (X or O) not used by B.
7.30. (a) The game is a draw.
(b) A wins by forcing B to complete the middle
row.
7.31. A wins by starting in the middle on the
bottom.
7.32. A wins by starting in the bottom row, second
column.
7.33. A wins by starting in the middle.
7.34. (a) A wins by starting in one of the four
central boxes.

444
ANSWERS TO SELECTED PROBLEMS
7.35. A wins by starting in the center.
7.36. A wins by placing two adjacent X's in the
interior of the third row.
7.37. A wins by starting as in this figure:
O
7.38. A wins by taking one box.
7.39. (a) and (b) A wins by taking the middle row.
7.40. (a) A wins by starting in the center.
(b) A wins. See the Hints and Solutions section.
7.41. (a) A wins by starting in the center of the
lower level.
(b) A wins by forcing B to complete three O's
in a row on the second level.
7.42. A wins by starting with an X in the middle,
with an O below it.
7.45. (a) A wins by starting with the fifth or sixth
cell.
(b) B wins.
7.46. (a) B wins. (b) B wins.
(c) A wins by starting in the center.
7.47. (a) B wins.
(b) A wins by starting with the two central cells.
7.48. (b) A wins by starting in the center and
playing by central symmetry until the opportunity
to win presents itself.
7.49. (a) B wins. (b) The game is a draw.
(c) B wins, 2 to 1.
7.50. (a) (i) A wins by starting in the center.
(ii) A wins easily by starting in one of the two
cells on the middle of the shorter diagonal. (He can
also win by starting on one of the end cells of the
shorter diagonal, but in this case the game is more
complicated.)
(iii) A wins by starting in the center cell.
(b) B wins on the n x n boards for n odd; and
A wins on the n x n boards for n even. However, we
do not know the winning strategy except on the
3x3 and 4x4 boards.
7.51. (a) A wins in all three cases.
(b) B wins in both cases.
7.52. (a) (i) B wins.
(ii) A wins by starting in the center and then
playing by central symmetry.
(iii) A wins by starting in the middle of the top
row and playing by vertical symmetry.
(iv) A wins.
(v) We think B wins.
(b) (i) A wins.
(ii) and (iii) B wins.
7.53. (i) If n is odd, A wins; if n is even, B wins.
(ii) If n is odd, B wins; if n is even, A wins.
7.54. (i) A wins. (ii) B wins.
7.55. (a) (i) A wins. (b) (i) B wins.
7.56. (i) A wins except when «= 1, 5, 9, or 13.
(ii) A wins except when « = 2, 3, or 6.
7.57. (a) (i) A wins by starting in the middle row.
(ii) B wins.
(b) (i) A wins by starting in the center.
(ii) B wins.
(c) (i) A wins by starting in the middle of the
second row. (ii) A wins by starting at the left
side of the second row.
(d) (i) B wins. (ii) B wins.
7.58. (i) B wins. (ii) B wins.
7.59. (a) (i) B wins. (ii) A wins.
(b) (i) A wins.
(c) (i) B wins.
7.60. (i) B wins.
(ii) A wins.
(ii) B wins.
7.61. (a) B wins.
(b) A wins by starting on the left.
(c) B wins for « = 2, 3, 6, 9, or 10; A wins in the
other cases. For « = 4 or 8, A starts by pushing on
the left side; for « = 5 or 7, A starts by pushing
in the center.
(d) A wins by pushing in the center.
7.63. (i) (a) Draw. (b) We think it is a draw.
Certainly B has at least a drawing strategy.
(ii) (a) Draw. (b) A wins. (c) A wins.

ANSWERS TO SELECTED PROBLEMS
445
7.64. (a) A wins by starting in the center and using
central symmetry until B completes the third vertex
of a square.
(b) B wins. (c) A wins. (d) B wins.
7.65. (i) (a) A wins by placing the checker in the
lower lefthand corner.
(b) B wins.
(c) If n is even, then B wins; if n is odd, then A
can win by proper placement of the checker.
(Proper placement depends on the parity of w.)
(ii) (a) A wins by placing the checker in a cell
next to the corner.
(b) B wins.
(c) If « is even, then B wins; if n is odd, A wins
by proper placement of the checker. (Proper
placement depends on the parity of w.)
7.66. A wins by placing the first checker in the exact
center of the table.
7.67. (a) B wins.
(b) A wins by drawing a loop starting and ending
at one dot so that the remaining two dots are
separated from each other.
7.68. (a) (i) The game is a draw. (ii) A wins.
(iii) A wins.
(b) No.
7.71 (a) The dwarfs win.
(c) The giant wins.
(b) The dwarfs win.
7.72. (a) Fred wins. (b) Fred wins.
7.75. Either in d-3 or f-2.
7.79. Play Id-le. Then don't allow your opponent
to isolate any more points.
7.82. (a) Move to C-4. (b) Move to C-4.
CHAPTER 8
Practice Problems
8.A 1. (a) 12; even.
(c) 0; even.
(b) 13; odd.
8.B 1. (a) Possible: 3, 5, 4, 2, 1, 3, 5, 4, 2, 1, 4,
5, 3, 4, 1, 2, 5, 3.
(b) Impossible. (c) Impossible.
Exercises
8.2. (a) 5. (b) 10. (c) 21. (d) 42.
(e) 1 + 2^ + 2^ + • • • + 2" - 1 = i(2" + 1-1).
(f) 2^ + 2^ + 2^ + • • • + 2" - 1 = i(2" +1-2).
8.3. (a) 4. (b) 6. (c) 8. (d) 11.
(e) M, = 14; Mg = 17; M, = 20; M,o = 24.
8.4. (a) and (d) See the figure below.
(a)
8.5. (b) See the figure below.
(d)
/
/
/
/
/
(b)
8.6. (a) 1 in all cases.
(b) (i) 22. (ii) 11. (iii) 6.
(c) (i) 2. (ii) 2. (iii) 1.
(d) (i) 30. (ii) 30. (iii) 15.
(e) None in all cases.
8.7. (a) (i) None. (iii) 1.
(b) (i) 6. (iii) 4.
(c) 1 in all cases.
(d) (i) 13. (iii) 7.
(e) 1 in all cases.
8.8. Here are the T and the boat (next page):

446
ANSWERS TO SELECTED PROBLEMS
8.9. (a) Yes.
(b) Not necessarily; it depends on the color of the
missing cell.
(c) No.
(d) See the Hints and Solutions section.
(e) A region with nine cells will suffice.
(f) Here are the two rectangles:
(m) 11, 32, 41, 44, or 53. Actually, we have
found solutions that would leave the single peg in
any of these holes except 32.
8.12. (a) 15 moves are needed.
(b) n(n + 2) moves are required.
(c) 120 moves are required.
(d) 34-33, 32-34, 31-32, 33-31, 43-33, 23-43,
13-23, 33-13, 53-33, 54-53, 34-54, 32-34,
33-32, 35-33, 45-35, 43-45, 23-43, 22-23,
32-22, 12-32, 11-12, 31-11, 33-31, 53-33,
55-53, 54-55, 34-54, 44-34, 43-44, 23-43,
21-23, 31-21, 33-31, 34-33, 32-34, 12-32,
13-12, 33-13, 53-33, 43-53, 23-43, 33-23,
35-33, 34-35, 32-34, 33-32.
8.15. It can be done in 12 moves, with one of the
cousins ending up in the foyer and the den vacant.
8.17. One solution is shown in this figure:
r
nr
(h) 35.
8.10. (a) The center cube or any cube lying in the
middle of an edge of the 3x3x3 cube.
8.11. (a) The pegs that started in 24, 42, 46, or 64.
(d) The statements are true of all of the holes
indicated by X in the figure below.
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
(e) Start by jumping 34-32.
(j) It can be done in four moves starting with
33-53-35-13-31.
(k) No.
G
C}
Xb
Y
O
Xb
R
/
R
/
W
/
w
back
G
G
//o
B
Y
X'o
R
/
R
/
/
W
/
w
back
Y Y
front
Lower level
H
w
/b
^
G
/^B
R
/
O
/
o
/
w
front
Upper level
R
W
Xo
B
Y
//g
R
/
O
/
/
o
/
w

ANSWERS TO SELECTED PROBLEMS
447
8.19. (a) Yes. Start sowing from cup number 3
(just before the bowl).
(b) No. (c) No. (d) No.
(e) Yes. Start sowing from cup number 3 (the
second cup before the bowl).
(f) No. (g) No. (h) No. (i) No.
CHAPTER 9
9.3. 4, 9, 10, ace, 3, 6, 8, 2, 5, 7.
9.6. First weigh three of the coins against three
other coins. This will limit the possibilities to three
coins. Then weigh one of these against another.
9.11. (a) Weigh one coin from the first sack, two
from the second sack, etc.
9.13. It can be done in fourteen reversals, with
each train ending up in its original order.
9.18. No.
9.19. No.
9.20. Yes: All people who believe that knowledge
is obtainable through reason would take a chance that
a bucket of paint might fall on their heads.
9.22. 499. Any way of doing it will require 499
moves.
9.24. 233.
9.25. 95 millimeters.
9.26. 74.
9.27. (a) 14. (b) 3.
9.29. The die on the right in the first figure can't
be either of the dice in the bottom figure unless a
number is repeated on the die.
9.30. There is such a set; it contains 50 numbers.
9.31. (a) One answer is to make knots at the 1 inch,
4 inch, and 6 inch marks on the string.
APPENDIX A
A.l. (a) X = 3 or -2.
(b) X = i
(c) x = l
(d)x:
9 + ^73
73
(e) X > -|.
(f) No solution.
(g) x = S,y= -3.
(h) X = 3, 3; = 4 and x = 4, y
(i) x= \,y= -3,z = 5.
-3.

448
ANSWERS TO SELECTED PROBLEMS
Appendix C Answers
PPS C.A
1. a) (h,h,h), (h,h,t), (h,t,h),(h,t,t), (t,h,h), (t,h,t),
(t,t,h),(t,t,t).
b) i) (h,h,h), (h,h,t), (h,t,h), (h,t,t)
ii) (h,h,h), (h,t,t), (t,h,h), (t,t,t)
iii) (h,h,h), (h,h,t), (h,t,h), (t,h,h)
3. outcome set one: (use r for right and w for
wrong)
a) {(r,r,r,r), (r,r,r,w), (r,r,w,r), (r,r,w,w), (r,w,r,r),
(r,w,r,w), (r,w,w,r), (r,w,w,w), (w,r,r,r), (w,r,r,w),
(w,r,w,r), (w,r,w,w), (w,w,r,r), (w,w,r,w), (w,w,w,r),
(w,w,w,w)}
b) i) (r,r,r,r)
ii) (r,r,r,w), (r,r,w,r), (r,w,r,r), (w,r,r,r)
iii) (r,r,w,w), (r,w,r,w), (r,w,w,r), (w,r,r,w),
(w,r,w,r), (w,w,r,r)
outcome set two: {4r, sriw, 2r2w, irsw, 4w}
b) i) 4r
li) 3rlw
iii) 2r2w
PPS C.B
1.
2.
a)
b)
0
7
8
PPS c.c
1. a)
b)
1
2
1
2
1
2
7
8
ito5
5toi
PPS CD
1. a) Equally likely outcomes: {(h,h,h), (h,h,t),
(h,t,h), (h,t,t), (t,h,h), (t,h,t), (t,t,h), (t,t,t)}.
b) o heads: |
1 head: |
2 heads: |
3 heads: |
PPS C.E
1.
^^ 21- 1
^>' 52 - 2
b) -^ = ^
^J 52 13
c) -^ = ^
^J 52 26
d) ^ = ^
^J 52 13
^J 52 13
PPS C.F
1. a) 4-510.4 = 800
b) 4.5-10.4.4 = 3200
PPS C.G
1. a) 5040
b) 40320
c) 56
d) 28
PPS C.H
1. a) 720
b) 30240
c) 3628800
d) 10
e) 1
PPS C.I
1. 479001600
3. 79
PPS C.J
1. a) 120
b) 120
c) 1
d) 1
e) 1
f) 252
PPS C.K
1
' 19958400
PPS C.L
1. a) (x + yY = x"* + 4x^y + 6x^y^ + 4xy^ + y^
b)
(x + y)' =x' +5x'y + \0x'y' -hlOx'y' +5xy' + y'
c) (x - yY =x' -3x'y + 3xy' - y'
3. o
PPS CM
PPS C.N
1. £, and £4 are independent.
PPS CO

ANSWERS TO SELECTED PROBLEMS
449
PPS C.P
C.12 i
2
9 •
1. a) i^r=iir
b) i-(il)'-4-(i|)^(i|) or
(il)^+4(i|)^(i|)4-6(i|)^(i|)^
c)
PPS C.Q
1 1_ (1)10 =1021
^' ^ V2/ 1024
PPS C.R
1. 5 to 1
PPS C.S
^' 13 ^52 52 52 ^ '--^ CCULb
PPS C.T
^' 8 ^ 8 ^ 8 ^ 8 2 '
$1.50
C.i a) 5 (including one pile with all the cards),
b) 2.
C.2
C.3
C.4
C.5
C.6
C.7
C.S
C.9
C.io
C.ii
78.
a) ^.
b)i|.
c) -^ = -^
'^^ 20 5 •
a) -^^ = .143>-!-.
4800 7
b) 0.
0 i
Jill selected 5 and 15
a) 60.
b) i.
1
120 •
1
30 •
24.
30.
a) 792.
b) 945.
C.13
C.14
a) 2592.
b) 72.
0 i
C.15 576.
C.16 a)
b)
26
10
.13.
0
26
10
isVpo
10 10
^ .000012.
.04.
C.17
C.18
C.19
C.20
C.21
C.22
C.23
C.24
C.25
C.26
2316.
a) 17576.
b) 3125-
a) 625.
b) 3471875.
c) 5557.
a) 780.
b) 3542650.
a) 4536.
b) 24917760.
c)i658.
35.
14.
a) ^
b) ^
3840.
a) 3612.
b) 3136.

450
ANSWERS TO SELECTFO PROBLEMS
C.27
C.28
C.29
c) 3472.
d) 2576.
e) 3696.
a) 55.
b) 170.
c) 50.
(*}
a) 319-
b) -^«.02
216
C.30 a) 10! = 3628800.
b) 2(9!) = 725760.
c) 9! = 362880.
d) 2(8!) = 80640.
6!
C.31 — = 360.
2
C.32 76.
C.33 83.
C.34 a) ^^« 6.3x10"^^
52
13
b)
4
\M
Ll«J
(39) fl3^
+
UJ {9)
(39) (13^
+
UJ lioj
r39ii
UJJ
4
\('A
Ll'u
(39) A3^
+
[2) [\2)
r39^ fia^
+
lU ll3J
M^
Ivj
c)
a
4
.005.
f'^l
Lv'J
f.39^
+■
UJ
fn^
• +
^13J
r^^ii
lojj
4
d) -L
■,t: in:
-4.3
52
13
13Yl3^ r4Yl3Yl3Y26
7 jl 6
2 jl 6 A 6 A 1
e)
.21.
4-3
[0
m^
("]■■
■<")
("]]
= 9.8
^•3
10
26] fl3Yl3
2r 6
(']
[2)
[f"l
LUj
f"l
Uj
h^f'1
lU l5J
f'1
I5J
f'11
UJJ
g)
h)
"AT
<■:)
f"l
UJ
M'']
UJUJ
f'^l
IJ
«.ll
.03
«.04.

ANSWERS TO SELECTED PROBLEMS
451
13Y131 (131 fl3
4 il 3
4 j I 1
f52]
UJ
iM'ii':
f52]
UJ
=^.15.
mm
e.')
~ 1A
k)
52
13
12 ¥40^ fl2Y40
4 Uh 5 8
.(")
f")
UJ
w*
(;^]
[:]
.34.
48
-7
1) >-v«.003
^52^
13
20
vl3
m)7-^« 1.2x10
^52
13
36
n) >-4 « .004
^52^
13
13
C.35 a) ^^=^2.0x10 "^
52
5
. 10(45)
c) Vn« 9.9x10 ^
^52^
40 _5
d) -^^ =« 1.5x10 ^
52]
13
48
V1; -4
e) / / ^ 2.4x10
^52^
52
5
4 Y 12 V 4
52
5
13Y41 \V
2 il2J " \ 1 .
^52
13V.5 A.^.5 il3
i4^-(ioy-4r;u4o
C.36 a) straight flush: o;
2V45
v2A 1
four of a kind: ^ ; \ ^ « .003
47
b) o.

452
ANSWERS TO SELECTED PROBLEMS
full house:
TKil
flush: o
straight: o
three of a kind:
::
21(9
1 [ 1
«.01
;n
.11
two pair:
+ 9
<UiW
one pair:
'Mm\i
.16
47
4Y3
1 A 1
no pair: o
b) straight flush:
four of a kind:
full house:
flush:
straight: ^
three of a kind: o
two pair: o
^.71
one pair:
no pair:
47
11
47
C.37 941
C.38
C.39
a) 2.
b) 9.
c) 44.
d) 265.
e) ^.
265
2293839
3628800
922
.033.
: 0.63.
C.40
4165
« 0.22.
C.41
C.42
C.43
C.44
C.45
a)
b)
a)
b)
c)
d)
1
6
a)
b)
0
5
36
36 \36
^«0.36.
6
1
8 •
1
4 •
1
7
1 — 1.
8 4
23
— « .47.
49
4
— « .29.
14
117
« .64.
182
d) stand pat.
a)
b)
0
22
« .45.
49
3
— «.21.
14
856
«.55.
2184
d) stand pat.
14105
(36)^
= 0.30.

ANSWERS TO SELECTED PROBLEMS
453
C.46 a) ^^.27.
v2; 3
b) -^ = -^A3.
7
C.47 a) ^^«.55.
b) -«.43.
7
5^ fs
c) ^ ^^ X - - ^ -43.
71 7
15
N^.^O dy
b)
c)
d)
C.49 1 -
- ,6 -32~-
f'l
ii) ^ = —-.31.
2^ 16
3
5'
2
5"
22
50'
365-364 •...•341
T, --57.
365""
(Note it is greater than —.)
2
C.50 a) I -
10
1
1048576
« 9.53x10
,10
, . 3 "l 59049
b) - = « .05.
,4 J 1048576
c)
109
262144
4.15x10
C.51 a) .684.
b) .967.
c) .999.
C.52 a) .0001.
b) .3024.
c) .6976.
C.53 a) 32.
b) One route ends in slot 1; five in slot 2;
ten in 3; ten in 4; five in 5; and one in 6.
1
c) -.
C.54 512.
^ 56
C.55 a) «.026.
2187
b)
2851
19683
.145.
, 8272
c) « .420.
19683
C.56 7 to 5.
C.57 a) — «097.
72
2
b) — «.074.
27
17
c) « -$.08.
216
C.58 -$.125 « -$.13.
C.59 a) Thomas.
b) Neil.
c) Rudolph.
d) Rudolph wins with probability
27
10
Thomas,, with probability —; and
27
Neil, with probability —.
27

454 ANSWERS TO SELECTED PROBLEMS
24
e) Rudolph's expectation is -$19— ^ -$19.89;
27
22
Thomas'is -$51— « -$51.81; and Neil's is
27
19
-$71— « -$71.71.
27
C.6o a) $3.
b) There is none; no matter what he pays the game is
favorable to him. (The expectation is infinite.)
C.6i
C.62
C.63
C.64
C.65
C.66
-$.29.
14-
42.
1680.
462.
a) 1.
b) 2.
c) 1.
d) 4-
e) 394.
C.67 The corner of G Street and Fourth Avenue.
C.68 60.
C.69 a) 15376.
b) 400000.
C.70 276.

Index
Pages on which definitions are given are in boldface type.
Additivity, 343
Age problems. See Problems, age
Algebra, 70-84
Algebraic techniques, 321-331
Alphametic, 157
Analyzing a game
frontal assault, 233-234
limiting factors, 234-239, 241
working backward, 230-232,
246-247
Ancient problems. See Problems,
ancient
"And." See Conjunction
Andersson, G., 263
Antecedent, 44
Appel, K., 198
Argument, 38, 55-60, 377
form of, 38, 56, 57, 58, 59
valid, 56-57, 58, 59-60
Aristotle, 38
Assumptions, 9, 12-13, 15-17
secondary, 14
Axiom, 137
Bachet, de Meziriac, 1,151
weight problem, 146, 151-153
Ball, W. W. R., 276
Base, 125, 146, 148-156, 165-168,
172
addition and multiplication,
153-155
conversion, 148-150
ten. See Decimal notation
two. See Binary system
Bergholt, E., 288
Biconditional statement, 45, 44-46,
48
Binary system, 151-152, 165-166,
246, 247-249
Binomial coefficient, 350
Binomial Theorem, 350, 350-351
Bipartite graph, 177, 402, 403
Birdcage, 372
Bishop, reflecting, 252
Bishop's move, 203-204, 207
Blackjack, 370
Boole, G., 38
Boss Puzzle. See Fifteen Puzzle
Bouton, C. L., 247
Boxing Tournament Problem, 2, 4
Branch, 17, 19, 176. See also Edge
Branch point, 17
Bridg-it, 224, 262, 269
Calendar, 144
Cancellation law, 128
Cardano, 337
Carroll, L., 37, 38, 317
Case analysis, 16-17
"Casting out nines," 125-127
Chance, 215-216, 218
Checkerboard games, 214, 215,
263-265, 266, 269-270
Chess, 213, 217, 221,251,252
Chevalier de Mere, 337
Chinese rings, 414
Chuckluck, 372
Chuquet, R, 95, 96
Circuit, 178
Eulerian, 178, 182-184, 186, 190,
192-193,203-205
Hamiltonian, 192-193, 194,
205-208, 401
Clavius, C, 95, 96
Coefficient, 350
Colored cubes, 275, 299-304,
311-312
Instant Insanity, 299
Coloring, 282-284, 285, 286-287,
298-299,312,404,416
of graphs, 200
of maps. See Map coloring
Combination, 348, 348-349
Common divisor, 116
Complement of an event, 340
Complement of a graph, 402, 403
Complete bipartite graph, 177, 403
Complete graph, 177, 403
Composite number, 106
Conclusion, 55-56, 417, 418
Conditional probability, 352-353,
370-371
Conditional statement, 44-46, 48
Congruence, 121, 120-131, 132,
134^137
Conjecture, 198, 202
Fermat's, 100-101, 104n, 202
Four Color, 198
Conjunction, 42-43, 48
Connected graph. See Graph,
connected
Connectives, 41
biconditional, 45, 4446, 48
binary, 41
conditional, 44, 4446, 48
conjunction, 42-43, 48
negation, 41, 42, 48
unary, 41
Consequent, 44
Contradiction, 9, 12, 13, 48
Contrapositive, 54
Conway, J. H. 265
Counting, 343-349, 366-370
Cram, 263-264
Craps, 338, 354-355, 362-363, 370
Cross-cram, 263
Crossing problems. See Problems,
crossing
Cryptarithmetic, 156-162, 169-172
"Cubby Hole" Principle, 402^03
Decanting, 117-118
Decimal notation, 125-126, 146,
163-165
Decimation, 313-315
Deficient number, 100
Definition, 137
Degree of a vertex, 180, 186, 199
even, 180, 199
in-degree, 190
odd,180-186
out-degree, 190
455

456
INDEX
Deja vu, 241, 252
De Morgan, A., 38
Dependent events, 352
Dependent experiments, 351
Dice, 337, 338, 353, 356, 361
Dicircuit, 190
Eulerian, 190
Dicube, 308
Digit, 147
Digital root, 125-127
Digraph. See Graph, directed
Dimensional consistency, 80
Diophantine equations, 103-104,
115, 117
linear, 115, 117, 119, 131-137
Diophantus, 70, 95, 96, 103, 104
Dipath, 191
Eulerian, 191
Directed graph, 190
Disjoint events, 343, 344
Disjunction, 43-44, 48
Dissection problems, 279-282,
306-307
Dividend, 120, 160
Divisibility, 104-105
properties of, 105
testsfor, 351-352, 395
transitivity of, 105
Division algorithm, 120
Divisor, 104, 111-113, 116, 120, 160
greatest common, 116, 128
number of, 112
Dominating set (of vertices), 211,
404
Dominoes, 205, 274, 281, 282-283,
308
Dots, 261-262, 268-269
Drawing strategy. See Strategy,
drawing
Dual map, 199
Dudeney, H. E., 102
Dwarfs and Giant, 267
Edge, 176
Edge-disjoint path, 187-189
Elimination, method of, 77-78,
326^327
End position problems, 253,
268-272
Entertainment value, 365
Entrapment, games of, 266-268
Equally likely outcomes, 339, 342
Eratosthenes, 107
Euclid, 100, 106, 119
Euclidean algorithm, 119
Euler, L., 173, 175
Euler diagrams, 417-418
Eulerian circuit. See Circuit,
Eulerian
Eulerian dicircuit, 190
Eulerian dipath, 191
Eulerian path. See Path, Eulerian
Even vertex. See Vertex, even
Event, 339
Eves, H, 70
Expectation, 362, 361-364, 372-373
Expected payoff, 361
Expected value, 362, 372-373
Experiment, 338
Extraneous solution, 83
Factor, 105
Factorial, 139, 334, 346
Factoring
integers, 83
quadratic polynomials, 325
Fair coin, 340, 363
Fair game, 363
Fair odds, 362
Favorable outcomes, 338
Favorable game, 363
Fermat, P., 100, 101, 104n, 337
"Last Theorem," 100, 104n, 202
Fibonacci sequence, 418
Fifteen Puzzle, 274-275, 288-299
sliding block puzzles, 311
Finiteness (of a game), 216, 218
Five Color Theorem, 199
Four Color Conjecture, 198
Four Color Theorem, 198, 199, 200,
202
Fox and Geese, 266-267
Fox and Hounds, 266, 270
Frege, F L. G., 38
Fundamental Theorem of
Arithmetic, 109-110
Gale, D., 262
Gale, game of, 262
Games
analysis of See Analyzing a game
bounded, 217
chance-free, 215-216, 218
checkerboard. See Checkerboard
games
entrapment, 266-268
finite, 216-217, 218
join of, 405
lengthof, 217, 218-221,413
matchstick. See Matchstick
games
misere form, 252, 254, 255
of perfect information, 216, 218
regular form, 254, 255
state of, 224-225, 228
two-person, 213, 218
variations of, 251-252
Gardner, M., 1, 213, 253, 264, 320
Gauss, C F, 122
Golomb,S., 281,308
Graph, 176, 175-212, 303-304
bipartite, 177, 402, 403
coloring, 200
complement of, 402, 403
complete, 177, 403
complete bipartite, 177, 403
connected, 179, 182, 185, 187,
190,193
directed, 190
nonplanar, 177
planar, 177
traceable, 178
Grasshopper, 252
Greatest common divisor, 116, 128,
130, 131-132
Greek Anthology. 70, 95
Grundy's Game, 256
Guthrie, F, 198
Haken, W, 198
Hamilton, W. R., 192
Hamiltonian circuit. See Circuit,
Hamiltonian
Hamiltonian path. See Path,
Hamiltonian
Hein, P, 223, 260, 262, 285
Hex, 223, 222-224, 262, 269
Hexapawn, 264, 269 270
Hexominoes, 308
Hi Q. See Peg Solitaire
Highway inspector's problem,
187-189
Hip, 264-265
Honest die, 340
Hunter, J. A., 157
"If and only if" See Biconditional
statement
"If, then." See Conditional
statement
Impossible event, 340
Incidence, 180
Independence, 351, 352, 353,
370-371
Independent events, 352, 353
Independent experiments, 351
Independent vertices, 212, 404
Inductive reasoning, 332
Inference problems. See Problems,
inference
Instant Insanity. See Colored cubes

INDEX
457
Integers, positive, iOO, 104
Intersection, 344
Inversion, 291, 292
Join of games, 405
Josephus Problem, 3 i 3
Journal of Recreational
Mathematics, 320
Kalah, 312
Kayles, 255
Khowarizmi, al-, 70
Knight's move, 194
Knight's tour, 194-195, 207
reentrant, 194, 195,207, 312
Konigsberg Bridge Problem,
173-174, 175, 181-182,
186-187,201-202
Labyrinth, 4-5
Ladies Diary, 102
Leibnitz, G. W, 38
Length of a game, 217
example of, 218-221,413
Linear equations, 322, 325-330
dependence of, 326n, 327n
inconsistent, 326n, 327n
systems of, 325-329
Linear inequality, 323
Listing order, 290-291, 294
Logic, 37
Logic problems. See Problems,
inference
Logical equivalence, 54
Logical implication, 54
Loop, 176
Lottery, 373
Loyd, S., 288, 289, 294
Luck. See Chance
MacMahon, PA., 311
MahavTra, 95, 96
Mancala, 312
Map
coloring, 198-201, 335, 418^20
dual, 199
planar, 198, 199
Matching problems. See Problems,
matching
Matchstick games
more than one pile, 214-215,
246-249, 255
one pile, 3, 20-22, 214, 218-221,
227-228, 230-232, 254, 256
Mathematical induction, 152, 153,
218-220,278,332-335,
418^20
Metrodorus, 1, 70,95
Misere form (of a game), 252, 254,
255
Modulus (modulo), 120, 121, 128
Monochromatic triangle, 403
Moore's Nim, 255
Multigraph, 176n
Multiple, 105
Multiplication Principle, 19-20, 42,
113,234,290-300,345,356
Necessary condition, 45
Negation, 41,42, 48
Network, 176, 176n. See also Graph
Nim, 214-215, 246-249, 268
even and odd positions, 248-249
Moore's, 255
Northcott's, 258
Wythoff's, 255
Node, 17, 176. See also Vertex
Northcott's Nim, 258
"Not." See Negation
Noughts and Crosses. See Tic Tac
Toe
Odd vertex. See Vertex, odd
Odds, 341, 338, 361, 362, 372-373
"Or." See Disjunction
exclusive sense, 43
inclusive sense, 43
Outcomes, 338
Oware, 312
Pacioli, 337
Pairing, 404
Pairing strategy, 250-251, 370
Pairwise disjoint, 344
Parity, 248, 287
in Fifteen Puzzle, 293-294,
295-296, 298
in Peg Solitaire, 287
of a permutation, 292
Pascal, 337
Paterson, M. S., 265
Path, 17, 178
dipath, 191
edge-disjoint, 187-189
Eulerian, 178, 179, 181-182, 185,
186, 187, 190, 191-192,
202-205
Hamiltonian, 193, 195, 196
Pawn move, 264
Peg Solitaire, 274, 285-288,
308-310
French version, 309
Hi Q, 274, 286-288, 308-309
Pentominoes, 281, 308, 415
Perfect information, 216, 218
Perfect numbers, 100
Permutations, 290-296, 298-299,
346, 345-348
even, 291, 293
odd,291, 293
Phillips, H., 38
Pigeolet, M., 147
"Pigeon Hole" Principle, 402^03
Place value, 147
Planar graph, 177
Planar map, 198
Point of view, 4
Poker, 370
Polycube, 308
Polyominoes, 281-282, 284, 308,
415
Position (of a game), 225, 224-227,
228, 404
drawing, 225-227
equivalence of, 235, 238
examples of, 236, 237, 238-240,
413
even and odd. See Nim
initial, 225, 226-227
losing, 225-227, 228
penultimate, 233
starting. See Position, initial
terminal, 225, 226
winning, 225-227, 228
examples of, 229, 230-232,
241-245,246-249
Positional notation, 147-148, 154,
156-157
Potential payoff, 361
Premise, 55-56
Presenting a solution, 11
Prime factorization, 109-110, 111
Prime number, 100,106-111, 137,
332-333
infinitude of, 106-107
relatively, 116
Probability, 338, 337-365
of an event, 339
Problems
age, 72, 74-75, 77-78, 87-88
ancient, 72, 95-96
counting, 366-370, 373-375
crossing, 175, 196-197,208-209
cryptarithmetic, 147, 157-162,
169-172
decanting, 101, 117-118
decimation, 313-315
dissection, 279-282, 306-307
inference, 2-3, 26-35, 38-39,
61-69
logic. See Problems, inference

458
INDEX
matching, 2-3, 26-28, 30, 32-35,
66-67
odds, 341,372
shunting, 3, 22-25, 35-36,
316^317
tracing, 174,202-205
truthful-liar, 38-39, 61-66
uniform motion, 72, '^8-79,
88-92 I
weighing, 94^95, 146,! 151-153,
168,315-316
well-posed, 12
work,72, 81-84, 92-93
Proof (nature oO, 137,202
Proposition. See Statement
Propositional variable, 41
Pythagoras, 100
Pythagorean Theorem, 279, 280,
306-307, 414
Quadratic equations, 324-325,
329-330
Quadratic formula, 83, 324-325
Queen's move, 211, 258
Quotient, 120, 160
Reentrant knight's tour, 194, 312
Relatively prime numbers, 116
Remainder, 120, 123-124, 125, 127,
351-352
Remainder class, 120-122
Repeated experiments, 356-361
Repeated trials, 356, 371-372
Retracing, 189,204-205
Rhind Papyrus, 95
Rook's move, 207, 257
Roulette, 340, 360, 362, 363
Ruma, 312
Shader, L., 373
Sheep and Wolves. See Fox and
Hounds
Shunting problems. See Problems,
shunting
Sieve of Eratosthenes, 107-108
Silverman, D. L., 262
Sim, 266,403, 413
Simmons, G. J., 266
Simplification, 22, 232-233
examples of, 22-25, 241, 246,
289-291
Simultaneous equations, 325-330
Sliding block puzzles. See Fifteen
Puzzle
Slither, 262-263, 269
Solving problems
point of view, 4
presenting solutions, 11
simplification, 22-25
techniques, 11-12
visual aids, 10
Soma, 284-285
Sphinx, 1, 156
Sprague-Grundy method, 229
Sprouts, 265, 413
State diagram, 228-229, 230
State of a game, 224-225
Statement, 39^0, 49
compound, 49
Strategy, 21-22, 215, 221, 225, 228
drawing, 215,218, 221,227
examples of 407^08, 408^09
pairing, 250-251,410
winning, 215, 218, 221, 222, 224,
226-227, 409
examples of, 227-228, 230-232,
245,250-251,366,368-369,
406,408^09,410,412
how to find, 218-221, 228-229,
230, 232-234
Subscripts, use of, 80
Substitution, method of, 77-78,
325-326
Sufficient condition, 45
Summation, 350
Summers, G., 38
Syllogism, 38, 317-318, 417^18
Symmetry, 237-238, 234-240, 252
central, 235, 408
diagonal, 236
as a game strategy, 250-251, 406,
408
horizontal, 235
as a limiting factor, 234-236,
238-240, 243, 246, 406, 408,
411,417
vertical, 235
Tac Tix, 260, 268
Tangrams, 280-281, 307-308
Tetrominoes, 281, 282, 308
Theorem, 137,202
Tic Tac Toe, 224
symmetries of, 234-236
three-dimensional, 260-261,
270-271
variations, 214, 238, 258-261,
270-271
Tower of Brahma, 274, 276-279,
305
Tower of Hanoi. See Tower of
Brahma
Traceable graph, 178. See also Path,
Eulerian
Tree diagram, 16-18, 19-20,230
examples of, 17-18, 20-22, 42,
112,218-222,355,357-359
Tricks, 71,74, 96-99, 141-143,
164-165, 165-168,205,314
card, 96-99, 141-142,314
domino, 205
number guessing, 71, 74, 96,
142-143, 164-165, 165-168
Tricube, 285, 308
Trominoes, 274, 281, 283-284
True odds, 340
Truth table, 42, 43, 44, 45, 54
Truth value, 39^0
Truthful-liar problems. See
Problems, truthful-liar
Unfavorable game, 363
Unfavorable outcomes, 338
Uniform motion problems. See
Problems, uniform motion
Unique Factorization Theorem,
109-110
Validity, 56, 55-57, 59-60, 418
Variables, 72-73, 80
propositional, 41
Variations of a game, 251-252
Vertex, 176
of degree one, 193
of degree two, 188, 193, 195
even, 180, 182, 186, 190
odd, 180, 181, 182, 185, 186, 187
Vertices, set of
dominating, 211, 404
independent, 212, 404
Visual aids, 10
Weighing problems. See Problems,
weighing
Weight, 339
Well-posed problem, 12
Winning strategy. See Strategy,
winning
Work problems. See Problems, work
Wylie, C. R., 38
WytholT's Nim, 255